Thursday, November 30, 2006

Friday's Test Topics

Here’s a list of topics for Friday’s test:

Precalculus Chapter 5 Test Topics:
Composite Argument Properties (memorize! – sine and cosine)
Double Argument Properties (sine and cosine)
Understanding the implications of the +/- in the half-argument properties
Composition of Ordinates – sketch the graph of the sum or product of two given functions
Discussion/Illustration of co-function and odd/even properties
Numeric proof/demonstration of Composite Argument properties
Transform an equation given in terms of cosine with a phase displacement to a linear combination of sine and cosine
Transform an equation given as a linear combination of sine and cosine to an equation in terms of cosine with a phase displacement
Use a composite argument property to prove a co-function property
Use given values to compute composite functions
Compare algebraic computed values to calculator values
Harmonic analysis
Composite Argument Identity
Application of Sum and Product Properties
Application of ½ -angle Properties

That’s it! The format is as expected – ½ calculator, ½ non-calculator. The calculator portion is significantly longer than the non-calculator portion, so budget your time accordingly. I’ll be in my classroom on Thursday after school and before school on Friday. I’ll also be available online later on Thursday evening. If you have specific questions Thursday night, email me!

See you in class!

"The best way to predict your future is to create it."
- Peter Drucker

Wednesday, November 29, 2006

Our Online Textbook

Ok, as promised, here's the link and class pass to our online textbook:

The website:

Our class pass: 2901-a2e9cc0b7

Have fun studying!

5-6 Double and Half Argument Properties

Hooray! I'm so excited that it's my turn to blog once again!

Now that we have learned all about all the other properties in the chapter, it is time to learn some more. This section focuses on the double and half argument properties.

You use the double argument properties when you are writing cos2x or cos(x+x), and the other trigonometric functions. Similarly, from these double argument properties, you can derive the half argument properties.

We use these properties for many different purposes, including finding the exact values without the use of our calculators....

Here are the properties in this section:

The Products and Squares of Cosine and Sine

We use thes when you have a sine and a cosine, or two cosines and two sines.

Example of when you use the products and squares of cosine and sine:

The Double Argument Properties (the important part)
note: we have to memorize the sine and cosine equations

Double Argument Property for Sine
sin 2A = 2sinAcosA

Double Argument Properties for Cosine
(yes, there are multiple equations for cosine)

Double Argument Properties for Tangent
(although we don't have to know it)

Example of when you use the double argument properties:

The Half Argument Properties for Sine, Cosine, and Tangent

Example of how to use the half argument property:

Find the exact value of sin 15° → think of it as half of 30

choose the positive because the angle is in the first quadrant, where sin is positive

So...basically, that's the lesson!

Here is a link just incase you want some extra help....

My Favorite Part...the personalization!

Always remember to check the flavors of the day!:

and also....

the best place EVER for cupcakes is Sprinkles of course!

a picture to show you the goodness...

the yummy holiday cupcakes! are next!!!

Tuesday, November 28, 2006

5.5 The Sum and Product Properties

Mkay, so one of the things we learned from Larry's lesson (below) was that sums and products of sinusoids with different periods create new sinusoidal patterns. We learned how to find a graph from a sum or product of sinusoids, and how to do the reverse, to derive these equations from the graph.

One thing we may not have realized that a graph can be both a product of two sinusoids and a sum of two sinusoids.

In 5.5, we find out how to go from a product of a graph to its sum, and vice versa, given one of the equations. (If we're not given an equation, we can figure it out easily given the graph using what we learned from Larry's lesson!) In doing this, we are introduced to the Sum and Product Properties!!!

Here they are:

And, yes, we should memorize them all.

But let's back up a couple of steps-- how did we get these equations in the first place, eh?

Now it's time to get out the good ol' graphing calculator.
In radian mode, por favor, graph the following:
1. y = cos(6x) + cos(4x)
2. y = 2cos(5x) x cos(x) (make this one bold)

Both graphs should be the same. This means that...

Besides the fact that the graph shows this to be true, we can also prove it by expanding the equation using composite argument properties, starting with the left side.

When we add these two equations, we get this:

and when we subtract them, we get this:

WHOA! kinda looks like...

... from earlier! Memorizing the general equations will make life a little bit easier, so that we can say:

instead of going through the long-winded derivation process

Now, let's try to do the reverse, to go from a sum to a product.

example dos:


I really had a hard time finding a link, but if this helps you (which it probably won't) then I encourage you to use it.

on a completely different note, I think for my personalization, I would like to contribute to the 'hip' and 'happening'-ness of this blogger site, If I may. ^_^

This isn't just math-- it's VOGUE math!!! okay, nevermind.


Monday, November 27, 2006

5-4: Composition of Ordinates and Harmonic Analysis

Well, after our long break, we get the strangest lesson that we have had so far...
I will try to help explain things a little bit better.

Basically, this lesson is about graphs that have sin/cos with another sin/cos, and what we have to do is to find the equation. There are two different types (which look different), which are either the sums of the two (cos/sin) or the products.

So...the equations' graphs look like this.
Yes, they are in degree mode ^_^.

Sum :
Blue - y = 3cos (x) + sin (4x)
Orange - y = 3cos (x)
Red - y = sin (4x)

Product :
Blue - y = 3cos(x)sin(4x)
Orange - y = 3cos (x)
Red - y = sin (4x)

Here are some steps to find the equation that were written down during class if you didn't see them.

Well, obviously the first step that you take is to find whether it is a sum equation or a product equation.

If it is the sum of two sinusoids, then...

1) Draw along the sinusoidal axis and determine the equation of the sinusoid you just drew. Usually, if I were to do this, it'd be the larger sinusoid (one with the larger period.)

2) Count the number of mini-cycles within one cycle of the sum of the sinusoids. (Note: this is to help find the one with the smaller period.

3) Determine the Amplitude of the mini-cycle (count the spaces between the larger sinusoid graph and the final graph to find the amplitude)

4) Determine if the cycle is sin or cos

5) Multiply the # of minicylces by the coefficient of x of the sinusoid found in #1.

If the graph is a product of two sinusoids then...

1) Draw the envelope graph.

2) Determine the equation for the envelope graph (the graph with the larger period.)

3) Count the number of mini-cycles within the envelope (the book has a great example of this, Pg 193)

4) Take this number and multiply it by the coefficient of x from #2

5) Determine if the mini cycles given are sin or cos.
If you cut the envelope in 1/2, then if you

1) Have symmetry, then both trig functions are the same (sinxsinx, cosxcosx.)
2) Don't have symmetry, then both trig functions are different (cos(x)sin(x), sin(x)cos(x) ).

6) Write your product equation.

7) Check on your grapher that you did correctly. (Mr. French did mention that these problems would most likely show up on the calculator portion =].)

Alright, now it's time for YOU to show your proficiency at these types of problems! Hurray?

Find the equation from the graph provided above.

Method :

First we figure out what type of equation it is, product or sum.
Obviously, it is a product equation because the amplitude is not always equal.

Next we draw the envelope graph, it should look something like this. The Blue line is what you should've drawn.

The envelope graph should be y = 2sin(x).

Alright, next we count the number of mini-cycles within the envelope, which is 10.
Then we multiply it by the coefficient of x in the larger-period sinusoid, which is 1. So it is still 10.

Now, we have to determine if the other sinusoid is a sin or cos graph. Well, to figure that out, we cut out 1/2 of the envelope and decide what it is. It isn't symmetric, therfore they are different functions. That makes it a cosine function.
From this information, we can now find the other function which is cos10x.

Therefore the answer is...

y = 2sin(x)cos(10x)

Amira...YOU ARE UP NEXT! for my personalization.
For those of you guys who have not yet realized what one of the coolest shows on TV is...
It is no other than Heroes on NBC!!!
It is on tonight at 9 P.M. and watch it if you have time, and if you do happen to miss it, you can always then go onto and then watch it there. Hope you guys now understand the lesson now, than before!!

Harmonic Analysis Practice

Hi everyone! I know today's class was a lot to think about - I've uploaded some worksheets to the class website that have proven to be helpful in exploring the concepts. The file name is Precalc_EXP_05.pdf. If you're confused, these should help!

Serious-minded people have few ideas. People with ideas are never serious. - Paul Valery

Sunday, November 19, 2006

5-3: Other Composite Argument Properties's almost Thanksgiving break!
This chapter is basically a bunch of formulas.

The Odd-Even Properties

Cosine and Secant are even functions.
cos(-x) = cos x
sec(-x) = sec x

Sine, Cosecant, Tangent, and Cotangent are odd functions.
sin(-x) = -sin x
csc(-x) = -csc x
tan(-x) = -tan x
cot(-x) = -cot x

You could also write
sin(-x) = -sin x as sin(x) = -sin(-x). This goes for the other odd functions as well.

It is easier to see why these properties are true by lookin
g at the graphs of some of the functions.

For cos(θ), you can see from the middle graph that if you replace θ with its opposite, you get the same y-value. The cosine function is symmetrical over the y-axis and therefore is an even function.

For sin(θ) and tan(θ), replacing θ with its opposite yields the opposite y value. These are odd functions and are symmetrical about the origin.

Cofunction Properties for Trigonometric Functions

θ = sin(90º-θ) and sin θ = cos(90º-θ)

cot θ = tan(90º-θ) and tan θ = cot(90º-θ)

csc θ = sec(90º-θ) and sec θ = csc(90º-θ)

The cofunction properties involve angles and their complementary angles, or angles that add up to 90º.

cos x = sin[(π/2)-x)] and sin x = cos[(π/2)-x)]

cot x = tan[(π/2)-x)] and tax x = cot[(π/2)-x)]

csc x = sec[(π/2)-x)] and sec x = csc[(π/2)-x)]

Composite Argument Properties for Cosine, Sine, and Tangent

cos (A - B) = cos A cos B + sin A sin B

cos (A + B) = cos A cos B - sin A sin B

sin (A - B) = sin A cos B - cos A sin B

sin (A + B) = sin A cos B + cos A sin B

tan (A - B) = (tan A - tan B)/
(1 + tan A tan B)

tan (A + B) = (tan A + tan B)/(1 - tan A tan B)

(I know that's a lot to remember, but these are really important!)

Here is Mr. French's device to help you remember them:

Cosine is the "weird" function that is counterintuitive, so think opposite. When you see a "-" on one side, you will need a "+" on the other side (and vice versa). Sine and tangent make sense. A "-" will translate to a "-" on the other side, as will a "+." You just have to remember that in the sine equation, sin comes first, not cos.

Example Problem - Solving an Equation Using the Composite Properties

Drawings, Explanations, and Practice

Personalization :)

How do you prove in three steps that a sheet of paper is a lazy dog?
1. A sheet of paper is an ink-lined plane.
2. An inclined plane is a slope up.
3. A slow pup is a lazy dog.

Edward, you're next!

Thursday, November 16, 2006

5-2 : Composite Argument and Linear Combination Properties

Section 5-2 : Compostie Argument and Linear Combination Properties
(This sounds is scary...but its not that scary.)

Property : The Linear Combinations of Cosine an dSine with Equal Periods :
B cos (x) + C sin (x) = A cos (x-D)
A = √(B squared + C squared)
D = Arctan (C/B)
*Note : Arctan is used instead of just inverse tan, because if inverse tan does not lie in the right quadrant you must add 180 degrees or pi depending on whether you are in degree mode or radian mode*

Property : Composite Argument Property for cos (A-B)
cos (A-B) = Cos A cos B + sin A sin B
How to solve for x (example problem)
problem : 4 cos (x) + 4 sin (x) = 2 [0 , 4π]
solving :
A = √(16 + 16) = √(32)
D = Arctan (4/4)
√(32) cos ( x - .785 ) = 2 divide by √(32) on both sides

cos ( x - .785 ) = 2/√(32) multiply by arccos on both sides

( x - .785 ) = arccos (2/√(32) ) perform arccos

x - .785 = +/- 1.209 + 2πn add .785 to both sides

x = +/- 1.209 + 2πn + .785
x = 1.209 + 2πn + .785
x = 1.994 , 8.227
x = -1.209 + 2πn + .785
x = -.424 , 5.859 , 12.142
*Note : -.424 is in red because it is less than 0 and does not answer the problem. We must add 2π to -.424 in order to find a solution.*
Answers :
1.994 , 8.227 , 5.859 , 12.142
Thats all folks! You never have to learn another thing for the rest of your lives! (I lied about that last part.)
Reminder to MADI (who needs to give me her camera) that she is the next blogger- person- thing-a-majig
This is a link to an interactive graph of the linear combination of cosine and sine with equal periods property. Check this out. It helps a lot.
Personalization : I want to thank the Canadian people who came out to wave -- with all five fingers -- for their hospitality. -- George W. Bush

Tuesday, November 14, 2006

Chapter 4 Review Solutions

Just a quick note - I've posted a document with the solutions to the Chapter 4 practice test on our class website:

See you tomorrow!

Wednesday's Test Topics

Here’s a list of topics for Wednesday’s test:

Precalculus Chapter 4 Test Topics:
Basic properties – Pythagorean, Reciprocal, Quotient
Equations versus Identities
Strategies and thought processes in solving Identities
Arcfunctions – graphical representations of arcsine, arccosine and arctangent
Arcfunctions versus inverse trig functions – criteria for selecting principal branches
Transformations and Identities
Determining solutions to trigonometric equations – graphically and algebraically
Domains and Ranges of Inverse Trigonometric Functions

That’s it! The format is as expected – ½ calculator, ½ Non-Calculator. I’ll be in my classroom on Tuesday after school, at the JPD rehearsal Tuesday evening and before school on Wednesday. I’ll also be available online later on Tuesday evening. If you have specific questions Tuesday night, email me!

See you in class!

The best years of your life are the ones in which you decide your problems are your own. You do not blame them on your mother, the ecology or the president. You realize that you control your own destiny.
-- Albert Ellis

Monday, November 13, 2006

4.6 Inverse Trigonometric Relation Graphs

Graphs and Principal Branches of Inverse Trigonometric Functions:
In order to graph the inverse of sin, tan or cos you must switch the x and y variables. The inverse graphs are not functions. So, you have to choose a part of the graph in order for it to pass the vertical line test and be a function. You do this by restricting the domain of the graphs so that only the principal branch shows. Also, be sure to include the origin (or make it as centrally located as possible) and if possible, make it positive. The Principal Branch must show all possible values.
The inverse graphs can be graphed using the parametric mode. For example to graph arctan you would type: X1= tan(T) and Y1= T. This will give you the entire graph.
The function mode only graphs the principal branch of the inverse trigonometric function.

~The inverse graph of sine and its principal branch:

Photobucket - Video and Image Hosting
Domain: [-1,1]
Range:[-π/2, π/2]
Quadrants: I,IV

~The inverse graph of cosine and its principal branch:

Quadrants: (I, II)

~The inverse graph of tan and its principal branch:

Photobucket - Video and Image Hosting

Domain: (∞, -∞) or all real numbers
range: (-Π/2, Π/2)
Quadrants: I, IV


y=cot-1 x
Domain: (∞, -∞)
range: (0,Π)
Quadrants: I, II

y=sec-1 x
Domain: x ≥ 1
range: (0,Π) and y ≠ Π/2
Quadrants: I, II

y=csc-1 x
Domain: x ≥ 1
range: (- Π/2, Π/2) and y ≠ 0
Quadrants: I, IV

Exact Values of Inverse Circular Functions:

Create a triangle from the given information to geometrically solve the problem.

For example: cos (sin^-1(3/5))

From looking at the sin^-1(3/5) we know that the opp is 3 and the hyp is 5. Therefore the adj is 4 because of the pythagorean theorem.

Photobucket - Video and Image Hosting

Now that we have our triangle and angle set up we can find the cos of the angle which is the adj of hyp. The answer is 4/5.

The Composite of a Function and Its Inverse Function:

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

This is only true if x is in the range of the outside function and in the domain of the inside function.

Example Problem:

tan^-1(tan(50º)) = ?

Method of Solution: Remember that f^-1(f(x)) = x if x is in the range of the outside function and in the domain of the inside function.
Answer: 50º is within the range of the outside function and the domain of the inside function, so the answer is 50º.

Link to an additional reference:

This shows excellent inverse trigonometric graphs.


JPD is just around the corner juniors!!! 5 days and counting. I would like you all to see the picture i have had to stare at all weekend while writing the descriptions for the silent auction items:
It's supposed to be Marilyn Monroe..

SAMMMM do not forget that you are next to post a blog!!

Good luck EVERYONE on your chapter 4 test on Wednesday (11/15)!!!

Hope my blog helps you when you are studying!!

circles in new lights

I tried to send this from word, but it didnt work at all. If I get it working, then this post will look better.

Today, we found a new light in the circle. The Ferris wheel was brought back, and we got Wednesday's quiz back. We went back to algebra and revived the ellipse.

There is your Ferris wheel. Isn't it beautiful? I made it in paint. Awesome...

Let’s say that the period for this function is 10 seconds. What would be the height at 5 seconds?

It would be the same as the initial height (1/2of a period)

How far to the left would the Ferris wheel be after 5 seconds?

The diameter....easy so far, right?


parametric equations to help with ellipses

first, to set your calculator:

Press mode and make sure you are in parametric (par) instead of function (func). Then make sure you are in degree mode.

Then to set your window: usually (at least for today), the window be 10x10. The t min and max should be 0 and 360 because one period is 360 degrees.


Actual equations

now we move on to real work.

Put the equations x=5cos (t) and y=7sin (t)

your graph should look like this

Notice in the equation that cosine corresponds with the x coordinates, and sin corresponds with the y coordinates.
Also, note that the ellipse moves from -5 to 5 on the x axis and from 7 to -7 on the y axis. The x axis is limited the way it is shown because the equation is 5cos (t). The 5 half of the x-axis diameter as the 7 in the sin equation shows half of the y axis diameter.
Now as you can see, the equations x=cos (t) and y=sin (t) produce a unit circle
lets try a different one. Let’s say for example....
we have an x axis diameter of 12, and a y axis diameter of 6.
What would the equation be?
you would need the radius instead of the diameter for both it would be x with a radius of 6, and the y axis with a diameter of 3.
What would the equations be then?
x=6cos (t)
y=3sin (t)
how to convert this weird form to a normal one
ex. the two equations:
x=5cos (t)
y=7sin (t)
first, square everything. make the equations as follows:
x^2=5^2 〖cos〗^2 (t)
y^2=5^2 〖sin〗^2 (t)

x^2/5^2 =〖cos〗^2 (t)
y^2/7^2 =〖sin〗^2 (t)

combine them
x^2/5^2 +y^2/7^2 =〖cos〗^2 (t)+〖sin〗^2 (t)=1
final equation:
x^2/5^2 +y^2/7^2 =1
now for shifting:
x=cos (t)+2
This would move the origin to the point (2,-3)
easy enough so far, right?
now....if you want part of an ellipse:

in order to restrict the part of the circle, do the divide by the limit
ex. half of circle....lets say that this is a unit circle
x=(cos^2(t))/(t≥0 and t≤180)
there you go......
you can also put the restriction under the "y" equation.....doesn't matter at all.

ok....question begins:
a. what is the equation for this in standard ellipse form
b. what would the 2 equations be if the origin was moved to (3, 4)
Lets say the X axis diameter is 8, and the y axis diameter is 10
a. The main way for this equation would be:
1= x^2/4^2 +y^2/5^2
b. if it is moved:
that it
reminder to Katie for next posting

here is another source to help you refresh your memory on ellipses

here is one of the greatest pictures ever
awesome pic


Tuesday, November 07, 2006

4-4: ArcSine, ArcTangent, Intervals, ArcLarrys

Alright folks, check your ego at the door because 4-4 is going to be a MAD STRUGGLE.

Actually, that's not true at all; it really that difficult unless you suck at math.

( And if that's the case, rejoice because this blog will solve all problems and remedy all conditions for the mathematically impaired among you)

We're going to be dealing with Arcsine, Arctangent, Weird Interval Things, and the amazingly intricate ArcLarry.

Please don't think less of me because I took notes with a purple pen. I swear, this is a completely isolated incident.

The theory behind Arccosine/Arcsine/Arctangent is essentially that each inverse function's ratio will yield multiple angles according to its period. If that made absolutely NO sense to you, then consider it this way: What we're trying to do is find multiple angles at which the ratio will be the same.

Here are the two fundamental ArcSine equations and an example of how they work:

And now....ArcTangent (it only has one equation because you merely need to add 180 n to find the next angle):

The following 3 charts are extremely useful in determining ArcSine/ArcCosine/ArcTangent. They indicate the quadrants in which you can find an angle that will produce the same ratio that the inverse function does.

Those Weird Interval Things (The mutated paranthesis basically denote whether x will be equal to the outside range of the interval. Check your book for a rad explanation)

AND FINALLY, the ArcLarry:

^ridiculously sweet link. I expect a gift basket or some show of appreciation for such a fantastic website.

"My country send me to United States to make movie-film. Please, come and see my film. If it not success, I will be execute. "


REMINDER TO NAVJIT. you're next. give 'em hell.