Hello, fellow classmates. My name is Paul, and I will be your scribe for today, and I will be writing about "odd" and "even" functions in an overview of Friday's class.

So our class today started with a review of yesterday's class (slide 2), which basically stated that:

In the formula y = af[b(x-c)]+d:

- If a > 1, the graph is stretched vertically

- If 0 < |a| <> 1, the graph is compressed horizontally

- The y-coordinates of f are multiplied by a

- If b > 1, the graph is compressed horizontally ("speeds up")

- If 0 < |b| < 1, the graph is stretched horizontally ("slows down")

- The x-coordinates of f are multiplied by (1/b)

Following the review, we spent some time practicing graphing functions, which can be seen on slide 3.

REMEMBER!

Stretches before translations!

Here's a quick rundown of the formula and what each variable does:

y = af[b(x-c)]+d

a -> Vertical stretch/compression

d -> Vertical shift

b -> Horizontal stretch/compression

c -> Horizontal shift

Note: Remember that you have to stretch the graph first, but it doesn't have to be both stretches first (abcd/abdc)! You can stretch the x axis and then shift it before stretching the y axis(adbc), the important part is that the stretch for each axis must happen before the shift for its respective axis. This was discussed on slide 5, where we wrote the possible orders you can apply transformations.

After the practice with drawing graphs, we were introduced to images, which is the graph after it has been shifted/stretched (slides 4 and 6).

Also, when talking about an image, the "formula" is:

(image) is the image of (original coordinates) under (function).

So an example would be:

(0,7) is the image of (-2,-2) under the function y=-3f[1/2(x-4)]+1

We then proceeded to learn all about the online bookmark service and website Del.icio.us.

I won't really go into detail about this, since it's pretty self-explanatory. However in the interest of thoroughness, I've made a quick image here:

And with that, we proceeded to our next topic of reflections (slide 7):

- Basically, a vertical reflection is a reflection along the x-axis which occurs when the y-coordinates of any function f(x) are multiplied by (-1).

- Conversely, a horizontal reflection is a reflection along the y-axis which occurs when the x-coordinates of any function f(x) are multiplied by (-1).

Slide 8 contains a picture of a reflected sine wave, neato.

Inverses are also introduced on slide 7:

As stated by the slide, the inverse of f(x) is f^-1(x) [f to the power (-1) then/applied to (x)]. Which looks like this (image follows):

Thank you, Paint, for that excellent demonstration.

And in BIG SCARY LETTERS, I WILL MAKE SURE YOU DO NOT FORGET THAT f^-1(x) IS NOT EQUAL TO [(1/f(x)]!!!

The truth is that (can you handle this?)... [f(x)]^-1 (note how the power is on the outside), is the one that is equal to [1/f(x)].

So in summary:

[f(x)]^-1 = [1/f(x)] √ Correct

f^-1(x) = [1/f(x)] X Wrong

Its also pronounced "eff inverse." And finally, f^-1(x) undoes what f(x) does.

Lastly, we started talking about "odd" and "even" functions (slides 10 and 9 respectively):

A function is "even" if it is symmetrical about the y-axis, which occurs only when f(x) = f(-x). If you look, a cosine curve is a nice example of an "even" function.

A function is "odd" if it is symmetrical about the origin, which occurs only when f(-x) = -f(x) (the negative of the whole function). If you look, a sine curve is a nice example of an "odd" function.

And that concludes my scribe post summarizing the topics we covered during our class last Friday. Now about the scribe...

Since some people actually want to be scribe (unlike myself who was hoping to be scribe for pi day), I hereby declare that benofschool has the honour of being the next scribe for Monday's class.

And don't forget to get a Delicious (mmm, delicious) account if you haven't already! Here's a direct link for you who are too lazy to type it (shame on you), or like convenient things.

Del.icio.us

So I bid you adieu and offer my sincerest apologies for the lateness of my post. Unfortunately, I had work Sunday and was busy the majority of Saturday. I did infact start this post on Friday, but did not manage to finish it until today. I am highly aware of the time I am posting this at, especially considering I had resolved to post it first thing on Friday.

Good night, and I'll see you tomorrow.

P.S. I would like to offer the following advice to the following scribes:

Blogger's draft system is not without its flaws, and I have had to rewrite the second paragraph (involving the overview of the a and b variables) numerous times because the draft system confuses my use of the arrow signs (<>) as parts of html tags and deletes everything inbetween. Thus, if you think you're not going to finish your post, just save yourself some hassle and put it in a notepad document instead of using the draft system.

## Friday, February 29, 2008

## Thursday, February 28, 2008

### Transformations

I am deeply sorry I posted so late, and I apologize to whomever was waiting for a scribe post. Unfortunately I didn't check the blog yesterday and hadn't realized that I was the next scribe as I was up working on a project for another class. Anyways, I'll recall as much as I can remember.

We started off the class waiting outside... but then he let us back in and let us write the second half of the test where we drew the graph of a function and wrote two functions for a graph. As you can see in the second slide, we went over the question just in case anyone u, anxieties, questions and any good jokes. In slide three we quickly went over what the answer was for the second question. He showed us a quick process to easily write the function. He found 'D' was 3 because the sinusoidal axis had moved UP 3 units. He found this by looking at the distance from crest to trough and found that distance was 6, so then he discovered D was 3. From that he also concluded that the amplitude was also 3. Finding the period, he saw that one full wave had occurred in the time of 'pi' so then he took B = 2pi/period, and found B = 2, as you can see in the slide. For Sine, there was no phase shift as the graph already started at the origin of the sinusoidal axis. For Cosine, he took the closest maximum value to the Y axis and found the phase shift is pi/4. So now we have the Functions. But don't do what I did and REMEMBER to put either y = 'function' or for whatever function its asking you to find (ex. f(x) = ...) You'll LOSE a WHOLE MARK as it is an EXPRESSION and not a FUNCTION.

We then looked at translations and how 'a' and 'b' respectively affect the function and it's position where 'a' affects the x coordinate and where 'b' affects the y coordinates as you can see in slide 4.

'A' is like the phase shift, and shifts the 'x' position of the function left or right, and b is like the sinusoidal axis, but really it just shifts the function up or down (depending on the sign).

We then looked at stretches. In f(x) = ... x is what is used to graph, so if in the case there is a 2 in the f(x) (ex. f(2x) or you can just look in the slides), the 2 would affect how the function looks. If the two happened to be in front of the whole function, the whole function would be multiplied by two. We then headed over to http://fooplot.com to see how they would get affected by an integer in front of the 'x'. The larger the integer, the smaller the function looks, because in a sense, it takes less time, as to where if it were an integer, that would mean it would take longer, thus stretching the graph out.

Unfortunately we didn't make it to Compressions.

It's probably no-use saying this at this time of the night but homework is exercise 8. Again, I'm truly sorry for posting so late, especially to the people who were looking for the scribe to post to look for some answers.

~Rence_Out

Oh yeah, Next scribe will be... *checks list*

DARTH PAUL! I CHOOSE YOU! QUICK ATTACK, NOW!

To simplify that unnecessary outburst of randomness,

We started off the class waiting outside... but then he let us back in and let us write the second half of the test where we drew the graph of a function and wrote two functions for a graph. As you can see in the second slide, we went over the question just in case anyone u, anxieties, questions and any good jokes. In slide three we quickly went over what the answer was for the second question. He showed us a quick process to easily write the function. He found 'D' was 3 because the sinusoidal axis had moved UP 3 units. He found this by looking at the distance from crest to trough and found that distance was 6, so then he discovered D was 3. From that he also concluded that the amplitude was also 3. Finding the period, he saw that one full wave had occurred in the time of 'pi' so then he took B = 2pi/period, and found B = 2, as you can see in the slide. For Sine, there was no phase shift as the graph already started at the origin of the sinusoidal axis. For Cosine, he took the closest maximum value to the Y axis and found the phase shift is pi/4. So now we have the Functions. But don't do what I did and REMEMBER to put either y = 'function' or for whatever function its asking you to find (ex. f(x) = ...) You'll LOSE a WHOLE MARK as it is an EXPRESSION and not a FUNCTION.

We then looked at translations and how 'a' and 'b' respectively affect the function and it's position where 'a' affects the x coordinate and where 'b' affects the y coordinates as you can see in slide 4.

'A' is like the phase shift, and shifts the 'x' position of the function left or right, and b is like the sinusoidal axis, but really it just shifts the function up or down (depending on the sign).

We then looked at stretches. In f(x) = ... x is what is used to graph, so if in the case there is a 2 in the f(x) (ex. f(2x) or you can just look in the slides), the 2 would affect how the function looks. If the two happened to be in front of the whole function, the whole function would be multiplied by two. We then headed over to http://fooplot.com to see how they would get affected by an integer in front of the 'x'. The larger the integer, the smaller the function looks, because in a sense, it takes less time, as to where if it were an integer, that would mean it would take longer, thus stretching the graph out.

Unfortunately we didn't make it to Compressions.

It's probably no-use saying this at this time of the night but homework is exercise 8. Again, I'm truly sorry for posting so late, especially to the people who were looking for the scribe to post to look for some answers.

~Rence_Out

Oh yeah, Next scribe will be... *checks list*

DARTH PAUL! I CHOOSE YOU! QUICK ATTACK, NOW!

To simplify that unnecessary outburst of randomness,

**Paul's**the next scribe.
Labels:
Lawrence,
Scribe,
Transformations

## Wednesday, February 27, 2008

### New Unit: Transformations

Hi everyone! Its finally my turn to scribe, and today we started a new unit. But first I must talk about what little happened during the morning class...

In the morning, we began by looking over the review questions that were on yesterday's slides. Mr. K told us not to worry about the second part of those review questions as much since we didn't really go over how to do them, but we did in the afternoon to some extent.

After we had finished briefly looking over those questions, we were given the first part of our Circular Functions test. Notice how I say "first part". The reason why I say this is because tomorrow morning, we will be getting a second part to the test, which should only take us about 5 minutes to do, according to Mr. K. The second part will consist of graphing work. That was all that happened in the morning class.

Onto the afternoon class, which we had a late start on because of Mr. K talking to the principal or something. Anyways, Mr. K first started off by showing off some trippy spiral themed pictures from www.flickr.com. Then he took it one step further by showing us his newly downloaded firefox add on, PicLens, which gave him the ability to look through all the pictures in a cool way. Its hard to explain for me so if you're interested in this, go to www.piclens.com!

Okay back to math! So, we started on a new unit called Transformations. The first thing we did was look at the graph for f(x) = x^2, which looks like this:

We then looked at what would happen when 2 was added to the function to make f(x) = x^2 + 2. The graph ended up looking like this:

As you can see, by adding 2, the parabola was shifted upwards by two units. We then looked back at the sine graph and did the same thing to it, except we subtracted 1, therefore causing the graph to shift downwards by one unit.

Now, if you're looking at slide 3 of today's class, you'll notice that we tried graphing y(x) = 2^x. Don't fret if you don't understand what this is. The reason why the graph looks like that is because since x is equal to a number, when 2 is to the exponent of that number, it will increase as so, making the graph look like that. To help you understand it better, here is a table that shows you a simple pattern of the relationship between exponents and a number, which in this case is 2:

See the pattern? Starting from exponent 4 all the way down to negative exponent 4, the values keep halving themselves. So if x is equal to 4, that would mean the y coordinate would be 16. If x was equal to -4, that would mean the y coordinate would be at 1/16.

Okay, so now that you know why it looks like that, what would the graph look like if I were to do make it f(x) = 2^(x+3)? Well, if you're still looking on that slide, then you'd see that the graph moved 3 units to the left. This is pretty much like the phase shift in those trig function graphs we did for the past few units. The only difference is that it isn't called the phase shift, since phase shift is only used when dealing with trig functions, instead, we call it the horizontal shift.

So in conclusion to this part of the afternoon class, we can say that we write these functions in the form of y = f(x-b) + a, where "b" is the horizontal shift and "a" is the vertical shift.

By the way, the weird shaped thing below the f(x) = 2^x graphs are called "piece wise functions". They are called this because of they are split into pieces. As you can see in that specific piece wise function, there are three parts to it, thus why its a piece wise function.

For the rest of the class, we were given questions to work on based on the things I just talked about. The first question, as shown in slide 6, asked us to to write f(x) = sin(x-2) + 5 in terms of g(x). Since f(x) is basically g(x) shifted 4 units to the left, the only possible answer was g(x) = sin(x+2) + 5. As you can see, the only thing that was changed was "b". To get +2, we basically just added 4 since f(x) is units to the left of g(x), meaning that we'd have to move right 4 units from f(x) to get to g(x). Simple, right?

Okay, so onto slide 9. Remember those review questions from yesterday's slide? Well, we learned how to solve those now. I'll use the same example as shown in slide 9, which was sin(2x) = 1/2. To make this easier to work with, you can Let θ = 2x and then work from there for now so that you've got sin θ = 1/2. Now, you should already know how to solve for this by now, and if you don't, look over the circular function notes again, mkay?

Yeah, so if you did it right, you should have gotten the answers in the black. If you're confused with the 13π/6 and the 7π/6, we just got those by adding 12π/6 which is equal to 2π to π/6 and the reason why we did this is because of the original 2x. Since its 2x, when you look at it on the graph, you'll see that there are two complete waves, meaning that there's going to be double the solutions. The same can be said if it was 4x instead of 2x. In the case of 4x, there'd be 8 solutions, and if it was 6x instead of 2x, there'd be 12 solutions. You see?

You think you're done with the question now eh? Well, you're not. Since you'd switched out 2x for θ, that means that you're going to have to put back the 2x. Then from there, you should be able to work with it, so if you did everything correct, you should get the answers in red. Voila, wasn't that oh so simple? :)

Um, so, I think I'm done now! Yay! Oh wait, just forgot to mention that tonight's homework is Exercise 7, questions 1-10, since you should have already finished 11-20 and if you haven't, you are a naughty child.

Now for the moment of truth, since I'm the last scribe in this cycle, that means I have a whole list to choose from now, ehehe. Therefore, I choose .....LAWRENCE.

Yeah, back to the beginning we go! Anyways, remember that there is part 2 of the Circular Functions test tomorrow, so good luck everyone!

~kristina

In the morning, we began by looking over the review questions that were on yesterday's slides. Mr. K told us not to worry about the second part of those review questions as much since we didn't really go over how to do them, but we did in the afternoon to some extent.

After we had finished briefly looking over those questions, we were given the first part of our Circular Functions test. Notice how I say "first part". The reason why I say this is because tomorrow morning, we will be getting a second part to the test, which should only take us about 5 minutes to do, according to Mr. K. The second part will consist of graphing work. That was all that happened in the morning class.

Onto the afternoon class, which we had a late start on because of Mr. K talking to the principal or something. Anyways, Mr. K first started off by showing off some trippy spiral themed pictures from www.flickr.com. Then he took it one step further by showing us his newly downloaded firefox add on, PicLens, which gave him the ability to look through all the pictures in a cool way. Its hard to explain for me so if you're interested in this, go to www.piclens.com!

Okay back to math! So, we started on a new unit called Transformations. The first thing we did was look at the graph for f(x) = x^2, which looks like this:

We then looked at what would happen when 2 was added to the function to make f(x) = x^2 + 2. The graph ended up looking like this:

As you can see, by adding 2, the parabola was shifted upwards by two units. We then looked back at the sine graph and did the same thing to it, except we subtracted 1, therefore causing the graph to shift downwards by one unit.

Now, if you're looking at slide 3 of today's class, you'll notice that we tried graphing y(x) = 2^x. Don't fret if you don't understand what this is. The reason why the graph looks like that is because since x is equal to a number, when 2 is to the exponent of that number, it will increase as so, making the graph look like that. To help you understand it better, here is a table that shows you a simple pattern of the relationship between exponents and a number, which in this case is 2:

See the pattern? Starting from exponent 4 all the way down to negative exponent 4, the values keep halving themselves. So if x is equal to 4, that would mean the y coordinate would be 16. If x was equal to -4, that would mean the y coordinate would be at 1/16.

Okay, so now that you know why it looks like that, what would the graph look like if I were to do make it f(x) = 2^(x+3)? Well, if you're still looking on that slide, then you'd see that the graph moved 3 units to the left. This is pretty much like the phase shift in those trig function graphs we did for the past few units. The only difference is that it isn't called the phase shift, since phase shift is only used when dealing with trig functions, instead, we call it the horizontal shift.

So in conclusion to this part of the afternoon class, we can say that we write these functions in the form of y = f(x-b) + a, where "b" is the horizontal shift and "a" is the vertical shift.

By the way, the weird shaped thing below the f(x) = 2^x graphs are called "piece wise functions". They are called this because of they are split into pieces. As you can see in that specific piece wise function, there are three parts to it, thus why its a piece wise function.

For the rest of the class, we were given questions to work on based on the things I just talked about. The first question, as shown in slide 6, asked us to to write f(x) = sin(x-2) + 5 in terms of g(x). Since f(x) is basically g(x) shifted 4 units to the left, the only possible answer was g(x) = sin(x+2) + 5. As you can see, the only thing that was changed was "b". To get +2, we basically just added 4 since f(x) is units to the left of g(x), meaning that we'd have to move right 4 units from f(x) to get to g(x). Simple, right?

Okay, so onto slide 9. Remember those review questions from yesterday's slide? Well, we learned how to solve those now. I'll use the same example as shown in slide 9, which was sin(2x) = 1/2. To make this easier to work with, you can Let θ = 2x and then work from there for now so that you've got sin θ = 1/2. Now, you should already know how to solve for this by now, and if you don't, look over the circular function notes again, mkay?

Yeah, so if you did it right, you should have gotten the answers in the black. If you're confused with the 13π/6 and the 7π/6, we just got those by adding 12π/6 which is equal to 2π to π/6 and the reason why we did this is because of the original 2x. Since its 2x, when you look at it on the graph, you'll see that there are two complete waves, meaning that there's going to be double the solutions. The same can be said if it was 4x instead of 2x. In the case of 4x, there'd be 8 solutions, and if it was 6x instead of 2x, there'd be 12 solutions. You see?

You think you're done with the question now eh? Well, you're not. Since you'd switched out 2x for θ, that means that you're going to have to put back the 2x. Then from there, you should be able to work with it, so if you did everything correct, you should get the answers in red. Voila, wasn't that oh so simple? :)

Um, so, I think I'm done now! Yay! Oh wait, just forgot to mention that tonight's homework is Exercise 7, questions 1-10, since you should have already finished 11-20 and if you haven't, you are a naughty child.

Now for the moment of truth, since I'm the last scribe in this cycle, that means I have a whole list to choose from now, ehehe. Therefore, I choose .....LAWRENCE.

Yeah, back to the beginning we go! Anyways, remember that there is part 2 of the Circular Functions test tomorrow, so good luck everyone!

~kristina

Labels:
kristina,
Scribe,
Transformations

## Tuesday, February 26, 2008

### Tuesday, February 26

My Turn to Scribe 2 ~ HOOOOOOOOOO!

It was my turn to scribe so...

At the start of class, Mr. K started by showing us a few websites with varying purposes. First he showed us skrbl.com which is basically an online whiteboard. Then he presented twitter.com which you can use to communicate with people but you have to sign up with an account.

After viewing the features of those sites, we finally got to buckle down and write our Pre-test. It was basically a scaled down version of a normal test. There were 2 multiple choice, 2 short answer, and 2 long answers. We were alloted 15 minutes. After our time limit was up, we assembled into groups of about 3 people. Each group would have to hand in the a paper with the best possible solutions. Following that we went over the answers to the test. Bell Rings*

ＴＨＥ END。

The next scribe is kristina.

It was my turn to scribe so...

At the start of class, Mr. K started by showing us a few websites with varying purposes. First he showed us skrbl.com which is basically an online whiteboard. Then he presented twitter.com which you can use to communicate with people but you have to sign up with an account.

After viewing the features of those sites, we finally got to buckle down and write our Pre-test. It was basically a scaled down version of a normal test. There were 2 multiple choice, 2 short answer, and 2 long answers. We were alloted 15 minutes. After our time limit was up, we assembled into groups of about 3 people. Each group would have to hand in the a paper with the best possible solutions. Following that we went over the answers to the test. Bell Rings*

ＴＨＥ END。

The next scribe is kristina.

Labels:
AnhThi,
Circular Functions,
Scribe

### Circular Functions BOB

Hello, hello again. I'm back to do some.. BOB-ing. Does anyone else find it strange that you call it BOB? Or maybe that's just me.. Anyways..

I think this unit, for someone who hasn't touched any math for a YEAR, was a pretty good unit to start with. I'm not saying it was easy, but it was simple enough that I was able to wrap my math-challenged mind around it, and that was more than awesome. I found that having the unit circle forever drawn in my head really helped me; to me, having that down is understanding everything else.

But what I'm really worrying about is the graphing, because I missed the classes that focused on it. But I am getting it, just reaally slowly, kind of like Jamie, but.. probably slower.

Soo, that's my BOB for Circular Functions. Good luck on the test tomorrow you guys. =)

I think this unit, for someone who hasn't touched any math for a YEAR, was a pretty good unit to start with. I'm not saying it was easy, but it was simple enough that I was able to wrap my math-challenged mind around it, and that was more than awesome. I found that having the unit circle forever drawn in my head really helped me; to me, having that down is understanding everything else.

But what I'm really worrying about is the graphing, because I missed the classes that focused on it. But I am getting it, just reaally slowly, kind of like Jamie, but.. probably slower.

Soo, that's my BOB for Circular Functions. Good luck on the test tomorrow you guys. =)

## Monday, February 25, 2008

### BOB, Circular Functions

Well the unit Circular Functions was a bit confusing for me in the beginning because of that whole converting degrees to radians it took me a week to understand it but now i get it. another thing that also confused me was the functions secant and co secant it confusted me because the words do not match haha but i get it now. Now the only thing that is troubling me is the graghing for some odd reason i get confused with the period part (B) in DABC another thing that i need to remmember is to put label my graph and also put arrows on my graph. The part that i liked about this unit is that you wont lose any marks for forgeting to put the untits for radian, because there are no units ahaha.

Well now i guess its time to study for the test on wednesday.

Richard

Well now i guess its time to study for the test on wednesday.

Richard

Labels:
BOB,
Circular Functions,
richard

### BOB on Circular Functions

Now that there's a test, I'm slightly nervous, but anxious to get it over with. This unit at first was quite simple and straight forward because converting degrees to radians in respect to proportions is quite easy.

The unit circle however, was really annoying because of all the memorizing we had to do. We had to memorize each point for every radian measurement on the unit circle, but everything in math is a pattern, and as soon as you have the pattern etched in memory then it should be no problem memorizing things such as the unit circle. I still sometimes have trouble with the points because I have yet to memorize it all. In time I will though. I guess I should draw out the complete unit circle a couple times without any outside help such as notes.

The next part I found worth mentioning is the trigonometry equations, we did these in grade 11, and I remember these quite well. I believe these equations were introduced to us in preparation for the sine and cosine waves that we are currently working on. Found from the equation f(x) = AsinB(x-C)+D or f(x) = AcosB(x-C)+D. Sure, it may look hard at first but with all great puzzles, you must put it together piece by piece. Plotting each value in order. Preferably in the order DABC. I found that plotting these were not too much of a hassle but I have yet to perfect it, due to simple mistakes. In time I will get wiser though.

Well that just about sums up my understandings on the unit of circular functions. All in all, this unit was a little rough around the edges, but still understandable.

Until next time!

-Francis

The unit circle however, was really annoying because of all the memorizing we had to do. We had to memorize each point for every radian measurement on the unit circle, but everything in math is a pattern, and as soon as you have the pattern etched in memory then it should be no problem memorizing things such as the unit circle. I still sometimes have trouble with the points because I have yet to memorize it all. In time I will though. I guess I should draw out the complete unit circle a couple times without any outside help such as notes.

The next part I found worth mentioning is the trigonometry equations, we did these in grade 11, and I remember these quite well. I believe these equations were introduced to us in preparation for the sine and cosine waves that we are currently working on. Found from the equation f(x) = AsinB(x-C)+D or f(x) = AcosB(x-C)+D. Sure, it may look hard at first but with all great puzzles, you must put it together piece by piece. Plotting each value in order. Preferably in the order DABC. I found that plotting these were not too much of a hassle but I have yet to perfect it, due to simple mistakes. In time I will get wiser though.

Well that just about sums up my understandings on the unit of circular functions. All in all, this unit was a little rough around the edges, but still understandable.

Until next time!

-Francis

Labels:
BOB,
Circular Functions,
Francis

### Jamie's First BOB

Hello universe, and hello there BOB...I think everyone's introduced. I see from previous posts, everyone knows BOB.

There's an upcoming test this week and the normal thing is to be nervous. But I'm not normal, I'm just something else..and if you're something else, you can have different feelings like anxiety and quivering in fear. But what way better to review than to talk about what was learned in the past month or so?

The first unit was about Circular Functions. Functions about Circles...that's just plain cool. Radical, really. and round.. I just like making the "R" sound....TORRRRRRRRRRRRRRRRRRRRRTILLA nailed it! But yes, no more tangents. This unit was a great introductory unit, and it made me feel better about myself because I felt so smart coming up with answers out of no where but that was the beginning. It was basically a review of grade 11 pre-cal, from where i just freshly sprung from last semester...We took "rules" like CAST and disregarded them-- REBELS!! and we took theories apart and properly determined the difference between an equation and a proportion.

BUT THERE IS ONE SLIGHT PROBLEM that might just lead to my demise in the near future. I am a SLOOOOOOW learner. Or as benofschool used to say, "as swift as a turtle"...actually slower than that. A sloth. I just have a problems with prioritizing [spelling?] thoughts in my head. I know the most random of things, like the lyrics to any Sweeney Todd song or how to talk in different accents, but I still can't grasp the whole idea of the unit circle and the DABC concept. Don't get me wrong, I know it and where the information comes from and understand it, it's just I can't register these values fast enough to finish a test!!!! URGH!!! I'm not making any excuses like a scapegoat, but I'm just saying I can do it...in about a century. The funny thing is, my magic number is not 6. It's more like 6, 295,048...I know I'm overexaggerating, but I'm getting the point across. I'm even trying a new thing [the idea came from the walkie talkie device on the blog] where I recorded myself singing the sincostan values in quadrant one and then making my way across all of the quadrants because I believe music is my TOOL. Yes, I try to be innovative you know....also modest, but I suppose I'm not doing a great job at that.

OKAY YES...Mr. K. did say to not MEMORIZE but REMEMBER...but sometimes I just the provided example to the idea of the paradox, FORGET TO REMEMBER. Here I was last year being told that the "VERTEX WAS EVERYTHING", but now it's more like "IF YOU DON'T KNOW THE UNIT CIRCLE, YOU WILL FAIL." How encouraging. and the PARAMETERS!!! for graphing these functions. But I promise to you world, BOB, and myself that I will master the art of circle-- THE ROUND SENSEI.

By the way, good luck on your tests everyone! Let's count down to it like it's another New Years DAY! Don't mind my being nutty, it's just I'm trying to think of the glass being half full right now. hahah

There's an upcoming test this week and the normal thing is to be nervous. But I'm not normal, I'm just something else..and if you're something else, you can have different feelings like anxiety and quivering in fear. But what way better to review than to talk about what was learned in the past month or so?

The first unit was about Circular Functions. Functions about Circles...that's just plain cool. Radical, really. and round.. I just like making the "R" sound....TORRRRRRRRRRRRRRRRRRRRRTILLA nailed it! But yes, no more tangents. This unit was a great introductory unit, and it made me feel better about myself because I felt so smart coming up with answers out of no where but that was the beginning. It was basically a review of grade 11 pre-cal, from where i just freshly sprung from last semester...We took "rules" like CAST and disregarded them-- REBELS!! and we took theories apart and properly determined the difference between an equation and a proportion.

BUT THERE IS ONE SLIGHT PROBLEM that might just lead to my demise in the near future. I am a SLOOOOOOW learner. Or as benofschool used to say, "as swift as a turtle"...actually slower than that. A sloth. I just have a problems with prioritizing [spelling?] thoughts in my head. I know the most random of things, like the lyrics to any Sweeney Todd song or how to talk in different accents, but I still can't grasp the whole idea of the unit circle and the DABC concept. Don't get me wrong, I know it and where the information comes from and understand it, it's just I can't register these values fast enough to finish a test!!!! URGH!!! I'm not making any excuses like a scapegoat, but I'm just saying I can do it...in about a century. The funny thing is, my magic number is not 6. It's more like 6, 295,048...I know I'm overexaggerating, but I'm getting the point across. I'm even trying a new thing [the idea came from the walkie talkie device on the blog] where I recorded myself singing the sincostan values in quadrant one and then making my way across all of the quadrants because I believe music is my TOOL. Yes, I try to be innovative you know....also modest, but I suppose I'm not doing a great job at that.

OKAY YES...Mr. K. did say to not MEMORIZE but REMEMBER...but sometimes I just the provided example to the idea of the paradox, FORGET TO REMEMBER. Here I was last year being told that the "VERTEX WAS EVERYTHING", but now it's more like "IF YOU DON'T KNOW THE UNIT CIRCLE, YOU WILL FAIL." How encouraging. and the PARAMETERS!!! for graphing these functions. But I promise to you world, BOB, and myself that I will master the art of circle-- THE ROUND SENSEI.

By the way, good luck on your tests everyone! Let's count down to it like it's another New Years DAY! Don't mind my being nutty, it's just I'm trying to think of the glass being half full right now. hahah

Labels:
BOB,
Circular Functions,
Jamie

### Monday, February 25

MY TURN TO SCRIBE ~ HOOOOOOO!

It was my turn to scribe so...

First of all, we started out by going over the test about maximum area of a triangle. So we have point P on the unit circle. B on the x-axis and it has to be a right angle triangle. What's the maximum area? Well we know that the area of a triangle is base x height divided by 2 and the base and height is equal to the x and y components of the triangle. We also know that the x and y components are equal to the sine and cosine of the central angle. We tested out the values and found that as the angle goes up the area would also increase but after

After that, we went over question N on the sheet. Turns it out it was not a typo and we actually had to factor out the 2 pi. This will give us a period of 1. After that we went over some more transformations and Ｉ a few important things we should know. For example, parameter B is not the period itself, but actually helps us calculate the period. If we substitute B in the formula:

A few minutes later, we were given a wave graphed on the Cartesian Plane and were asked to write the equation of the graph in sine and cosine. Turns out there are INFINITE ways of writing the equation so... there's nothing to complain about there and plus we will only be required to only write one or two (max) equations on tests and the provincial exam.

Somewhere in the class, we looked at the infamous tangent function. Finally after a few of us were dying to see it, it was revealed. The tangent function is kinda special. It appears to be like x³ but IS NOT. The curvature is slighty different. Another feature of the wonderful tangent functions is asymptotes. Asymptotes are areas where tangent cannot exist. Since tangent is a trigonometrical function it has to do with triangles. At certain angles, tangent does not form a triangle therefore tangent cannot exist. Tangent can also be infinitely large unlike sine and cosine which are confined to -1~1.

That pretty much wraps up the class. Until next time.

It was my turn to scribe so...

First of all, we started out by going over the test about maximum area of a triangle. So we have point P on the unit circle. B on the x-axis and it has to be a right angle triangle. What's the maximum area? Well we know that the area of a triangle is base x height divided by 2 and the base and height is equal to the x and y components of the triangle. We also know that the x and y components are equal to the sine and cosine of the central angle. We tested out the values and found that as the angle goes up the area would also increase but after

**π**/4 radians the area would go down again. That means the sine and cosine of**π**/4 radians will give up the largest area.After that, we went over question N on the sheet. Turns it out it was not a typo and we actually had to factor out the 2 pi. This will give us a period of 1. After that we went over some more transformations and Ｉ a few important things we should know. For example, parameter B is not the period itself, but actually helps us calculate the period. If we substitute B in the formula:

**period = 2π/b.**Another important tip we should know is if we are given parameter A = -1/2 and we are asked for the amplitude, then the answer would be 1/2 and not -1/2. This is because the amplitude is the distance from the sinusoidal axis to the maximum or minimum points. Therefore, distance cannot be negative because this would destroy all logic. In other words, read your questions carefully, if it asks for amplitude, then discard the negative sign but if it asks for the value of parameter A then you include the value of A.A few minutes later, we were given a wave graphed on the Cartesian Plane and were asked to write the equation of the graph in sine and cosine. Turns out there are INFINITE ways of writing the equation so... there's nothing to complain about there and plus we will only be required to only write one or two (max) equations on tests and the provincial exam.

Somewhere in the class, we looked at the infamous tangent function. Finally after a few of us were dying to see it, it was revealed. The tangent function is kinda special. It appears to be like x³ but IS NOT. The curvature is slighty different. Another feature of the wonderful tangent functions is asymptotes. Asymptotes are areas where tangent cannot exist. Since tangent is a trigonometrical function it has to do with triangles. At certain angles, tangent does not form a triangle therefore tangent cannot exist. Tangent can also be infinitely large unlike sine and cosine which are confined to -1~1.

That pretty much wraps up the class. Until next time.

Labels:
AnhThi,
Circular Functions,
Scribe

## Sunday, February 24, 2008

### BOB; Circular Functions

So knowing there's a test this week. I'll be BOBing.. while bobbing my head to music. Okay, lame, I know.

To the point. We started the semester without even introducing ourselves and our names which was a definite change from other teacher's ice breaking activities. And of we started, learning about Radians and what not, converting from degrees, a measurement we've been using in pre-cal for how long now, and finding our measurement was 'x' number pi over 'x' number. At first I had a hard time grasping the concept but later became comfortable after studying it for a while.

Later on, we learned the Unit Circle and all the radian measurements around the circle. Yet another item to remember, and the words of consequence from Mr. K when we questioned the idea of failing to memorize such a thing and he put it out plainly "You'll fail." And that's how the Pi pie crumbles... or is it how the cookie crumbles. Okay whatever, I'll stop with the mediocre attempts at what's called a joke. After much practice, I became comfortable with the Unit circle. Then finding the arc length was something else as well as we found that no matter how stretched out from the origin, it still shares the same proportionality.

The waves I found I was having a little trouble but had an easier time understanding than the whole Unit Circle memorization task. The only trouble I really had was with the Period ('B') but after Ben showed me how to do it it became clearer. I'll definitely study for this bad boy... Lol

G'niight all and good luck on the Test. Remember, the apple doesn't fall from the tree... wait what does that have to do with anything...?

-Lawrence

To the point. We started the semester without even introducing ourselves and our names which was a definite change from other teacher's ice breaking activities. And of we started, learning about Radians and what not, converting from degrees, a measurement we've been using in pre-cal for how long now, and finding our measurement was 'x' number pi over 'x' number. At first I had a hard time grasping the concept but later became comfortable after studying it for a while.

Later on, we learned the Unit Circle and all the radian measurements around the circle. Yet another item to remember, and the words of consequence from Mr. K when we questioned the idea of failing to memorize such a thing and he put it out plainly "You'll fail." And that's how the Pi pie crumbles... or is it how the cookie crumbles. Okay whatever, I'll stop with the mediocre attempts at what's called a joke. After much practice, I became comfortable with the Unit circle. Then finding the arc length was something else as well as we found that no matter how stretched out from the origin, it still shares the same proportionality.

The waves I found I was having a little trouble but had an easier time understanding than the whole Unit Circle memorization task. The only trouble I really had was with the Period ('B') but after Ben showed me how to do it it became clearer. I'll definitely study for this bad boy... Lol

G'niight all and good luck on the Test. Remember, the apple doesn't fall from the tree... wait what does that have to do with anything...?

-Lawrence

Labels:
BOB,
Circular Functions,
Lawrence

### My BOB for Circular Functions

The test is coming up soon so I thought the time was ripe for a BOB. Our very first unit is circular functions. At first, it was relatively easy. Nothing too difficult, I would say it was exactly like Grade 11 but units in radians instead of degrees. Took me a while to adjust to radians since we have been using degrees for most of our lives but as Mr. K always proclaims "FRACTIONS ARE OUR FRIENDS". Casting away CAST took zero effort whatsoever. As so, the memorization of the unit circle. Since it always has to be a root of 1, 2 or 3 over 2. Converting radians to degrees and vice versa was cakewalk since setting up a proportion is easy for me. All thanks to my Grade 8 Math teacher: MR. TRAN!

As things progressed on, we were exposed to more things that I found familiar from Grade 11. For example, solving for trigonometry functions of X was one of them. The only difference is that the angles are in radians now. Another thing was sine and cosine waves and their transformations.

I liked a few things about this unit. The fact that radians have no units was a load off my mind (in case I forget to write the unit on a test and lose a 1/2 mark). Solving for exact values of stuff was also enjoyable.

That brings my BOB to an end. Good night and Good Luck!

As things progressed on, we were exposed to more things that I found familiar from Grade 11. For example, solving for trigonometry functions of X was one of them. The only difference is that the angles are in radians now. Another thing was sine and cosine waves and their transformations.

I liked a few things about this unit. The fact that radians have no units was a load off my mind (in case I forget to write the unit on a test and lose a 1/2 mark). Solving for exact values of stuff was also enjoyable.

That brings my BOB to an end. Good night and Good Luck!

### BOB: Circular Functions

Hello everyone, since the test for circular functions is supposed to be this week, I'm here doing my BOB.

Okay, so during the beginning of this unit, when we were just introduced to converting degrees to radians and whatnot, I found that this was really easy. Sure, it was a bit confusing at first since I wasn't used to working with radians but it eventually got better as the unit progressed. Now I actually like working in radians better than degrees!

One part I had trouble with AT FIRST was learning the unit circle. As soon as I saw all those numbers on that unit circle, I panicked. What made it even worse was when Mr. K told us that we had to memorize EVERYTHING that was on the darn thing, and at that point, I was ready to cry. It didn't help much when Roxanne asked him what would happen if we didn't memorize the thing by the time he told us, to which he answered "You'll fail." But now that I know the cursed thing that caused me so much pain, I feel much better.

Now onto the graphing bits. The basic graphing wasn't so bad, but the added stuff, such as DABC, was. Sure, DABC was fun to say, but it was hard to understand at first since the last part of that class was rushed and it was so much to absorb in just those last 2 minutes we had.

Um, yeah, that's all I've got to say for this unit. It had its hard and easy moments, and overall, it was pretty good. That is all.

Okay, so during the beginning of this unit, when we were just introduced to converting degrees to radians and whatnot, I found that this was really easy. Sure, it was a bit confusing at first since I wasn't used to working with radians but it eventually got better as the unit progressed. Now I actually like working in radians better than degrees!

One part I had trouble with AT FIRST was learning the unit circle. As soon as I saw all those numbers on that unit circle, I panicked. What made it even worse was when Mr. K told us that we had to memorize EVERYTHING that was on the darn thing, and at that point, I was ready to cry. It didn't help much when Roxanne asked him what would happen if we didn't memorize the thing by the time he told us, to which he answered "You'll fail." But now that I know the cursed thing that caused me so much pain, I feel much better.

Now onto the graphing bits. The basic graphing wasn't so bad, but the added stuff, such as DABC, was. Sure, DABC was fun to say, but it was hard to understand at first since the last part of that class was rushed and it was so much to absorb in just those last 2 minutes we had.

Um, yeah, that's all I've got to say for this unit. It had its hard and easy moments, and overall, it was pretty good. That is all.

Labels:
BOB,
Circular Functions,
kristina

### BOB: Circular Functions

Since our test will be sometime this week, I decided to make my BOB entry before I forget. =)

In the beginning of this unit, I wasn't so comfortable converting degrees to radians. Up and till this unit I've always been using degrees but later in this unit, it became easy. This unit wasn't as bad as I thought it would be because from last year's precal I had trouble with circular functions. Now, I understand how to figure out certain word problems and convert degrees into radians and vice versa. I also found that finding the sine, cosine, and tangent of angles are pretty easy and it's probably because of the mini quizzes we had. But finding cosecant, secant, and cotangent can be a bit confusing at times and may take up a bit of time for me to figure out. What helps me a lot is to look for patterns to solve different kinds of questions which makes it much easier to do.

On the other hand, considering the fact that I missed a couple classes last week, I was a bit confused with the graphing we had to do. At first I was like, "what is this?" But after a few explanations and practicing doing questions I got the hang of it. Remembering DABC actually helped me a lot to do the problems. Yet, I'm not totally comfortable with graphing because it probably takes me more time than others in the class.

In conclusion, this unit was "OK" and I hope I do well in the first test as well as everyone else. With that said, goodluck to everyone!

PS..were we taught the graph for tangent when I wasn't here? Because in excercise 6, I believe there is a question about it and I didn't know how to do it?! Help?

In the beginning of this unit, I wasn't so comfortable converting degrees to radians. Up and till this unit I've always been using degrees but later in this unit, it became easy. This unit wasn't as bad as I thought it would be because from last year's precal I had trouble with circular functions. Now, I understand how to figure out certain word problems and convert degrees into radians and vice versa. I also found that finding the sine, cosine, and tangent of angles are pretty easy and it's probably because of the mini quizzes we had. But finding cosecant, secant, and cotangent can be a bit confusing at times and may take up a bit of time for me to figure out. What helps me a lot is to look for patterns to solve different kinds of questions which makes it much easier to do.

On the other hand, considering the fact that I missed a couple classes last week, I was a bit confused with the graphing we had to do. At first I was like, "what is this?" But after a few explanations and practicing doing questions I got the hang of it. Remembering DABC actually helped me a lot to do the problems. Yet, I'm not totally comfortable with graphing because it probably takes me more time than others in the class.

In conclusion, this unit was "OK" and I hope I do well in the first test as well as everyone else. With that said, goodluck to everyone!

PS..were we taught the graph for tangent when I wasn't here? Because in excercise 6, I believe there is a question about it and I didn't know how to do it?! Help?

### bob for Circular functions

Since our test is approaching for this unit I thought it was a good time to bob!

I was able to understand most things in this unit but my "muddiest point" would be the graphing part. I need still need some more practice with graphing those kinds of questions where all the DABC changes are there. I'm really comfortable with the A,B,C,D transformations but just not all together.

The part of this unit I really liked was probably those kinds of questions where you just have to find the exact values and simplify. At first that was kind of hard since we've all been accustomed to using degrees. I remember converting all the radians to degrees for each question, which was really time consuming. As I learned the exact values on the unit circle, I've found those kinds of questions way easier and didn't need to put it to degrees.

There's my bob for this unit. Good luck to everyone on the test this week?!

I was able to understand most things in this unit but my "muddiest point" would be the graphing part. I need still need some more practice with graphing those kinds of questions where all the DABC changes are there. I'm really comfortable with the A,B,C,D transformations but just not all together.

The part of this unit I really liked was probably those kinds of questions where you just have to find the exact values and simplify. At first that was kind of hard since we've all been accustomed to using degrees. I remember converting all the radians to degrees for each question, which was really time consuming. As I learned the exact values on the unit circle, I've found those kinds of questions way easier and didn't need to put it to degrees.

There's my bob for this unit. Good luck to everyone on the test this week?!

### BOB.1 [Circular Functions]

Well now that I'm back from my Kenora basketball trip, and once again have computer access I thought it would be an oppourtune moment to do my BOB post for this unit. Alas, I find myself here doing exactly that. BOBbing!

Well I'll start with the bad, and work my way towards the good, seeing as how it also goes like that in terms of chronological order from the beginning of the unit to the end :)

The hardest part of this unit was definitely the very beginning. Now this had alot to do with the fact that I missed the first two days of class, having been sick with the flu (>_<). So by the time I had returned to the class on Wednesday (that being the third day of class) I had basically no idea of what was going on. However, with some help from my classmates, and a couple nights spent awake much later then I should have been, I managed to figure my way around the first couple days of work, and catch up to everyone else.

On the flip side, the easiest part of the unit, has probably been the memorization, of the patterns in the unit circle (which eventually lead to the memorization of the unit circles basic values for sine, cosine, tangent, etc.) I found this rather easy, because all I had to do, was draw the unit circle over the weekend, without referring to my notes. How simple is that? So after drawing the unit circle, probably 7 or 8 times over the weekend, taking about 3-5 mins each time (with that time frame getting smaller as I went along,) I managed to get a fairly good grasp of basic angles locations on the unit circle, and the trigonometric function values of those said angles.

Overall this unit wasn't very difficult to understand or grasp (at least for me.) I'd say the hardest parts of it all were related to the memorization of certain things (like what A,B,C,D mean in graphing f(x) = AsinB(x-C) + D). I also found it somewhat difficult at times to remember what to do when adding radicals or similar things that we learned in previous grades, and I haven't had to use in awhile. Besides that I found much of what we did within my abilities to understand and put to use in the exercises we've been assigned.

Well I think that about sums it up for this BOB. I hope everyone enjoyed their long weekend and I also hope everyone enjoys the rest of today :) I shall see everyone on Monday!

Ciao!

~Justus

Well I'll start with the bad, and work my way towards the good, seeing as how it also goes like that in terms of chronological order from the beginning of the unit to the end :)

The hardest part of this unit was definitely the very beginning. Now this had alot to do with the fact that I missed the first two days of class, having been sick with the flu (>_<). So by the time I had returned to the class on Wednesday (that being the third day of class) I had basically no idea of what was going on. However, with some help from my classmates, and a couple nights spent awake much later then I should have been, I managed to figure my way around the first couple days of work, and catch up to everyone else.

On the flip side, the easiest part of the unit, has probably been the memorization, of the patterns in the unit circle (which eventually lead to the memorization of the unit circles basic values for sine, cosine, tangent, etc.) I found this rather easy, because all I had to do, was draw the unit circle over the weekend, without referring to my notes. How simple is that? So after drawing the unit circle, probably 7 or 8 times over the weekend, taking about 3-5 mins each time (with that time frame getting smaller as I went along,) I managed to get a fairly good grasp of basic angles locations on the unit circle, and the trigonometric function values of those said angles.

Overall this unit wasn't very difficult to understand or grasp (at least for me.) I'd say the hardest parts of it all were related to the memorization of certain things (like what A,B,C,D mean in graphing f(x) = AsinB(x-C) + D). I also found it somewhat difficult at times to remember what to do when adding radicals or similar things that we learned in previous grades, and I haven't had to use in awhile. Besides that I found much of what we did within my abilities to understand and put to use in the exercises we've been assigned.

Well I think that about sums it up for this BOB. I hope everyone enjoyed their long weekend and I also hope everyone enjoys the rest of today :) I shall see everyone on Monday!

Ciao!

~Justus

Labels:
BOB,
Circular Functions,
Justus

## Friday, February 22, 2008

### BOB Version 1: Circular Functions

Nope, not done blogging for today. Just doing my BOB…

And so, Circular Functions, the unit, put a lot of emphasis on the unit circle. Memorizing (no, not memorize, but remember!) the values using mnemonics was the muddiest thing that I didn’t like about this unit. But remembering the unit circle did help me out as we progress through the course. Also just as equally as muddy was the group work assignment, The Unit Circle Triangle. I didn’t like that assignment ‘cause of its “challengingness.”

I remember in gr.11 we were studying how to graph trigonometric functions and Mr.K exposed us to the equation f(x) = AsinB(x-C) + D. He taught us what A, B, and D did to the sine graph but told us that we’ll learn what B does in gr.12 precal. Now, after over a year of waiting, I found out what B meant and that became my moment of clarity. B stretches/shrinks the graph horizontally.

I also liked how Mr.K did his 1st class on the 1st day of school because I remembered it as inspiring.

I also liked the helpful advice Mr.K said to one of his students: Remember folks, Mr.K says, “If you don’t memorized the unit circle, you will fail.”

By the way, does anyone know how to do Exercise 2 #15?

If f (x) = 2x + 3, find k so that f(k+2) = k + f(k).

Answer key says:

2(k+2) + 3 = k + 2k + 3

2k + 7 = 3k + 3

4 = k

...which doesn't make sense to me.

Also, one of our questions in our homework asked us to graph the tangent function. We weren't taught that.

And so, Circular Functions, the unit, put a lot of emphasis on the unit circle. Memorizing (no, not memorize, but remember!) the values using mnemonics was the muddiest thing that I didn’t like about this unit. But remembering the unit circle did help me out as we progress through the course. Also just as equally as muddy was the group work assignment, The Unit Circle Triangle. I didn’t like that assignment ‘cause of its “challengingness.”

I remember in gr.11 we were studying how to graph trigonometric functions and Mr.K exposed us to the equation f(x) = AsinB(x-C) + D. He taught us what A, B, and D did to the sine graph but told us that we’ll learn what B does in gr.12 precal. Now, after over a year of waiting, I found out what B meant and that became my moment of clarity. B stretches/shrinks the graph horizontally.

I also liked how Mr.K did his 1st class on the 1st day of school because I remembered it as inspiring.

I also liked the helpful advice Mr.K said to one of his students: Remember folks, Mr.K says, “If you don’t memorized the unit circle, you will fail.”

By the way, does anyone know how to do Exercise 2 #15?

If f (x) = 2x + 3, find k so that f(k+2) = k + f(k).

Answer key says:

2(k+2) + 3 = k + 2k + 3

2k + 7 = 3k + 3

4 = k

...which doesn't make sense to me.

Also, one of our questions in our homework asked us to graph the tangent function. We weren't taught that.

### Wrapping Up and Concluding Circular Functions

3 assignments, 2 periods, 1 class. This is PC40SW08 (Pre-Calculus 40S Winter 2008).

As of now, Mr.K is still out-of-province because he was selected by The Council at the Conference last week to help write a textbook, but will return from his quest hopefully by Monday. And so, we had the same substitute as previous class'.

Period 1:

Our first assignment was Quiz 2 on Circular Functions in which we were not allowed to use our calculator. It was not a hand-in because the substitute says "according to the instructions Mr.K gave me, it did not say to hand anything in." So we assumed that we are not to hand that in. The class worked quietly during that quiz.

After finishing Quiz 2 on Circular Functions, we were assigned Graphing Trigonometric Functions. We were also told not to use a calculator when doing that assignment, but were allowed to use the calculator to check our answers. That is for homework, folks.

Period 5:

In our afternoon session, we were told to continue working on this morning's pre-calculus assignments and that Exercise 7: Translations #11-20 is also for homework.

HOMEWORK:

* Graphing Trigonometric Functions

* Exercise 7: Translations #11-20

* Circular Functions Test on Tuesday 26 or Wednesday 27

Next scribe is AnhThi.

That is all.

As of now, Mr.K is still out-of-province because he was selected by The Council at the Conference last week to help write a textbook, but will return from his quest hopefully by Monday. And so, we had the same substitute as previous class'.

Period 1:

Our first assignment was Quiz 2 on Circular Functions in which we were not allowed to use our calculator. It was not a hand-in because the substitute says "according to the instructions Mr.K gave me, it did not say to hand anything in." So we assumed that we are not to hand that in. The class worked quietly during that quiz.

After finishing Quiz 2 on Circular Functions, we were assigned Graphing Trigonometric Functions. We were also told not to use a calculator when doing that assignment, but were allowed to use the calculator to check our answers. That is for homework, folks.

Period 5:

In our afternoon session, we were told to continue working on this morning's pre-calculus assignments and that Exercise 7: Translations #11-20 is also for homework.

HOMEWORK:

* Graphing Trigonometric Functions

* Exercise 7: Translations #11-20

* Circular Functions Test on Tuesday 26 or Wednesday 27

Next scribe is AnhThi.

That is all.

## Wednesday, February 20, 2008

Hey everybody, Paul here, doing my scribe post (since I got "voluntold" to). I hope 8pm isnt too late for a scribe post, but I guess you guys don't really need it all that soon anyway as we didn't do a lesson.

So yeah, today in Pre-Calculus 40S we had a substitute (Mr. Rekrus?), and he gave us an assignment per Kuropatwa's orders. The assignment was on circular functions and it went something like this:

You have a unit circle, and on this circle you have a point P, which is along the circumfrence of the circle and in Quadrant I. You also have point A, which is at the center of the circle (thus making it the origin), and a point B, which is positive along the X axis. If triangle APB is a right angle triangle, find the value of P that creates the largest area of the triangle.

Since this was a group assignment, the class was split into three groups. There was also supposed to be a time limit of 20 minutes, but the question utterly stumped the groups for most of the class, so the time limit was disregarded. Some groups finished the question before the end of class, some spent the entire class trying to solve it and some groups made little paper cranes.

And that was what we did today in Pre-Calculus.

Today's homework is Exercise 6, Questions 1-20. I suggest you atleast try them since we do have a test coming up.

As for the next scribe, I have chosen Zeph since he asked so nicely.

Good night and farewell.

So yeah, today in Pre-Calculus 40S we had a substitute (Mr. Rekrus?), and he gave us an assignment per Kuropatwa's orders. The assignment was on circular functions and it went something like this:

You have a unit circle, and on this circle you have a point P, which is along the circumfrence of the circle and in Quadrant I. You also have point A, which is at the center of the circle (thus making it the origin), and a point B, which is positive along the X axis. If triangle APB is a right angle triangle, find the value of P that creates the largest area of the triangle.

Since this was a group assignment, the class was split into three groups. There was also supposed to be a time limit of 20 minutes, but the question utterly stumped the groups for most of the class, so the time limit was disregarded. Some groups finished the question before the end of class, some spent the entire class trying to solve it and some groups made little paper cranes.

And that was what we did today in Pre-Calculus.

Today's homework is Exercise 6, Questions 1-20. I suggest you atleast try them since we do have a test coming up.

As for the next scribe, I have chosen Zeph since he asked so nicely.

Good night and farewell.

Labels:
Circular Functions,
Paul,
Scribe

## Tuesday, February 19, 2008

### Graphing Sine and Cosine Functions

Hello, Francis here with your daily school scribe. Now I'll introduce you to the beginning of class. Let's bring it back.

Mr. Kuropatwa firstly showed us a new feature available on our blog. It's a translation feature that can translate this blog into one of many different languages. Brought to you by Google.

Mr. Kuropatwa also stated that the circular functions test will be sometime in the middle of next week. Which means that it would be a smart idea to start on your BOB posts.

Starting tomarrow there will be a substitute teacher and same goes for Thursday, because Mr. Kuropatwa won't be showing up for class on those days.

All of these functions that we graphed today are based on patterns and as said by Mr. Kuropatwa "Mathematics is the science of patterns". Okay now with today's lesson which was completely made up of graphing those sine and cosine functions, that are ever so interesting. As seen on slide 1 we started off class by graphing the sine function: y = sinx - 1 and on slide 2 the cosine function: y = 2cosx, then we stepped up to a slightly harder function found on slide 3 and I suggest you see how all that turned out, so check out that slide show. When graphing any of these functions you should remember a simple pattern: "1,2,3,4" where 1 is the maximum point, while 2 is the average point(between the max. and min. points) and 3 is the minimum point and 4 is the average once again. Remember when graphing one of these functions you should put arrows at the end, pointing in the direction the function is about to head to next, so if it's at the max point, and it has nowhere to go but down, you should end the line with an arrow pointing downward. Also remember not to put sharp points when you change direction from up to down or down to up, put curves, because if you put sharp peaks, then be prepared to lose your precious marks.

The general equation for these functions is: f(x) = Acos(x-C) + D or f(x) = Asin(x-C) + D.

When graphing these functions from these equations remember the word: DABC. This work or phrase is like the BEDMAS of graphing. It's the order in which you should always graph.

Now I will expose what these letters actually mean.

Firstly with D: is the vertical movement of the curve itself, it determines the sinusoidal axis(the axis between the max. points and min. points, its the average value of the curve.

Second is A: This is the amplitude it determines the max. and min. points of the curve on either side of the sinusoidal axis. The graph is stretched if the value of A is: |A| > 1 and its compressed if the value of A is: 0 < |A| < 1.

Third is B: this determines the period of the graph with an equation found on slide 18. If the value of B is negative then the graph curve will start with the pattern being 1,2,3,4. Where 1 is the min. point and 2 is the average point, and 3 is the max. point and lastly 4 being the average point.

Last is C: This shifts the curve left of right (horizontally). This is also called the "Phase Shift". If the value of C is negative(C <> 0) the curve shifts to the right.

For the sake of summarizing I will say this:

D - Lay down the sinusoidal axis.

A - Stretch of the amplitude.

B - Label the x-axis.

C - Look at the scale and slide the curve accordingly.

All of the details are found on slides 17-19. I suggest you check it out.

The sine and cosine curves are quite similar with the difference being that when at zero, cosine = 1 and sine = 0. But if the sine curve is slid to the right by pi/2, then at 0 sine = 1. With this, you can state that cosx = sin(x +(pi/2)) or sinx = cos(x-(pi/2)).

Well thats the 411 of today's class, hope it was helpful. The next helpful person to be voluntold(forced to do it for free, hence "volunteer that is told") is Paul.

That's me, the one and only, Francis! Until next time, I'm out.

Mr. Kuropatwa firstly showed us a new feature available on our blog. It's a translation feature that can translate this blog into one of many different languages. Brought to you by Google.

Mr. Kuropatwa also stated that the circular functions test will be sometime in the middle of next week. Which means that it would be a smart idea to start on your BOB posts.

Starting tomarrow there will be a substitute teacher and same goes for Thursday, because Mr. Kuropatwa won't be showing up for class on those days.

All of these functions that we graphed today are based on patterns and as said by Mr. Kuropatwa "Mathematics is the science of patterns". Okay now with today's lesson which was completely made up of graphing those sine and cosine functions, that are ever so interesting. As seen on slide 1 we started off class by graphing the sine function: y = sinx - 1 and on slide 2 the cosine function: y = 2cosx, then we stepped up to a slightly harder function found on slide 3 and I suggest you see how all that turned out, so check out that slide show. When graphing any of these functions you should remember a simple pattern: "1,2,3,4" where 1 is the maximum point, while 2 is the average point(between the max. and min. points) and 3 is the minimum point and 4 is the average once again. Remember when graphing one of these functions you should put arrows at the end, pointing in the direction the function is about to head to next, so if it's at the max point, and it has nowhere to go but down, you should end the line with an arrow pointing downward. Also remember not to put sharp points when you change direction from up to down or down to up, put curves, because if you put sharp peaks, then be prepared to lose your precious marks.

The general equation for these functions is: f(x) = Acos(x-C) + D or f(x) = Asin(x-C) + D.

When graphing these functions from these equations remember the word: DABC. This work or phrase is like the BEDMAS of graphing. It's the order in which you should always graph.

Now I will expose what these letters actually mean.

Firstly with D: is the vertical movement of the curve itself, it determines the sinusoidal axis(the axis between the max. points and min. points, its the average value of the curve.

Second is A: This is the amplitude it determines the max. and min. points of the curve on either side of the sinusoidal axis. The graph is stretched if the value of A is: |A| > 1 and its compressed if the value of A is: 0 < |A| < 1.

Third is B: this determines the period of the graph with an equation found on slide 18. If the value of B is negative then the graph curve will start with the pattern being 1,2,3,4. Where 1 is the min. point and 2 is the average point, and 3 is the max. point and lastly 4 being the average point.

Last is C: This shifts the curve left of right (horizontally). This is also called the "Phase Shift". If the value of C is negative(C <> 0) the curve shifts to the right.

For the sake of summarizing I will say this:

D - Lay down the sinusoidal axis.

A - Stretch of the amplitude.

B - Label the x-axis.

C - Look at the scale and slide the curve accordingly.

All of the details are found on slides 17-19. I suggest you check it out.

The sine and cosine curves are quite similar with the difference being that when at zero, cosine = 1 and sine = 0. But if the sine curve is slid to the right by pi/2, then at 0 sine = 1. With this, you can state that cosx = sin(x +(pi/2)) or sinx = cos(x-(pi/2)).

Well thats the 411 of today's class, hope it was helpful. The next helpful person to be voluntold(forced to do it for free, hence "volunteer that is told") is Paul.

That's me, the one and only, Francis! Until next time, I'm out.

Labels:
Circular Functions,
Francis,
Scribe

### BOB for Circular Functions

Hello everybody,

I believe the test on Circular Functions is this week so I decided to BOB ahead of time. This is our first unit of the course and I am doing great so far. The thing that really helped me in this unit is that activity that Mr.K got us to do and it was the one with drawing the unit circle and getting us to memorize those certain sine/cosine/tangent/cosecant/secant/cotangent of the special radian values. One thing I would like to do before the test is a little more practice. It should just give me a firm hold on the unit before the test.

Well that was my BOB and until our next text or my next scribe (which ever comes first) see ya and good luck on your tests.

I believe the test on Circular Functions is this week so I decided to BOB ahead of time. This is our first unit of the course and I am doing great so far. The thing that really helped me in this unit is that activity that Mr.K got us to do and it was the one with drawing the unit circle and getting us to memorize those certain sine/cosine/tangent/cosecant/secant/cotangent of the special radian values. One thing I would like to do before the test is a little more practice. It should just give me a firm hold on the unit before the test.

Well that was my BOB and until our next text or my next scribe (which ever comes first) see ya and good luck on your tests.

Labels:
benofschool,
BOB,
Circular Functions

## Saturday, February 16, 2008

### Circular Functions (Feb. 15, 2008)

Hello everyone it's Roxanne here and I am your scribe for today or shall i say tonight. I know it's pretty late but I just got home and I decided to do this before I forget.

We first started off with reviewing two questions that we should know how to do already. The questions should be on slide 2 if anyone needs to go back and check. For the second question, where it says cos(pi/2-pi/6), remember that cosine is a function and not a value to distribute. Such as the question 2(x+7), where it will equal 2x+14.

On to slide 3, we looked at the sine graph. It shows that if you look at the black wave the origin is zero aka sine θ=0, and as it moves to sine π/2 = 1 the curve goes up by one. Then it goes back on the x axis as it changes to sine π. As it changes to sine 3π/2 = -1, the curve is now at the bottom at -1 on the x axis. Finally, as it changes to 2π, it curve moves back on the x axis where it equals zero. It then repeats itself over and over again. REMEMBER: 1, 2, 3, 4! Therefore, if you change the input of the sine function such as sine x+1, the line would be higher than the orginial which is the black curve. If you change it to sine x-2, it would be below the black curve.

After we looked at slide 4 which shows us the the curve of cosine. But before we started talking about the graph, one of the students suggested that you could look at the the curve as two half circles. One would be on the top and one would be at the bottom. However, this is not true because in fact the curve of cosine behaves differently compared to the sine curve.

In our graphing calculator we put Y1= sin(x) and Y2=√(1-(x-π/2)². It then showed us the the cosine graph with almost, not quite a half circle on the top curve. Therefore, it proves that the curves are NOT two half circles.

We also used our graphing calculator to graph the cosine curve. Remember when using the graphing calculator go to zoom 4 aka the Friendly window. The friendly window does not create distortion such as zoom 6 where it takes the shape of a square rather than a rectangle. Mr. K also said that there are 95 pixels on the x axis and 63 pixels on the y axis. He explained that if subtract 1 from either of them, you get 47 on the x axis and 31 on the y axis. On calculator, it should show for the Xmin=-4.7 and the Xmax=4.7. It should also show Ymin= -3.1 and Ymax= 3.1. These are also the values that we calculated after you subtract by one and divided by two. The reason why you subtract one is because both the 95th pixel and 63rd pixel is counted as the origin.

Well I suppose that is all! Well at least I hope so because I am quite tired and I want to go to bed! So.. see ya later! OH ! I almost forgot, next scribe will be .......... the one and only FRANCIS! (: alrighty, bye folks!

We first started off with reviewing two questions that we should know how to do already. The questions should be on slide 2 if anyone needs to go back and check. For the second question, where it says cos(pi/2-pi/6), remember that cosine is a function and not a value to distribute. Such as the question 2(x+7), where it will equal 2x+14.

On to slide 3, we looked at the sine graph. It shows that if you look at the black wave the origin is zero aka sine θ=0, and as it moves to sine π/2 = 1 the curve goes up by one. Then it goes back on the x axis as it changes to sine π. As it changes to sine 3π/2 = -1, the curve is now at the bottom at -1 on the x axis. Finally, as it changes to 2π, it curve moves back on the x axis where it equals zero. It then repeats itself over and over again. REMEMBER: 1, 2, 3, 4! Therefore, if you change the input of the sine function such as sine x+1, the line would be higher than the orginial which is the black curve. If you change it to sine x-2, it would be below the black curve.

After we looked at slide 4 which shows us the the curve of cosine. But before we started talking about the graph, one of the students suggested that you could look at the the curve as two half circles. One would be on the top and one would be at the bottom. However, this is not true because in fact the curve of cosine behaves differently compared to the sine curve.

In our graphing calculator we put Y1= sin(x) and Y2=√(1-(x-π/2)². It then showed us the the cosine graph with almost, not quite a half circle on the top curve. Therefore, it proves that the curves are NOT two half circles.

We also used our graphing calculator to graph the cosine curve. Remember when using the graphing calculator go to zoom 4 aka the Friendly window. The friendly window does not create distortion such as zoom 6 where it takes the shape of a square rather than a rectangle. Mr. K also said that there are 95 pixels on the x axis and 63 pixels on the y axis. He explained that if subtract 1 from either of them, you get 47 on the x axis and 31 on the y axis. On calculator, it should show for the Xmin=-4.7 and the Xmax=4.7. It should also show Ymin= -3.1 and Ymax= 3.1. These are also the values that we calculated after you subtract by one and divided by two. The reason why you subtract one is because both the 95th pixel and 63rd pixel is counted as the origin.

Well I suppose that is all! Well at least I hope so because I am quite tired and I want to go to bed! So.. see ya later! OH ! I almost forgot, next scribe will be .......... the one and only FRANCIS! (: alrighty, bye folks!

Labels:
Circular Functions,
Roxanne,
Scribe

## Friday, February 15, 2008

### Today's Slides: February 15

Here they are ...

Here are the answers to your homework assignment:

Working with the Unit Circle Exercises - Answers

1. (i) 7π/3

(ii) 1

(iii) I

(iv) (1/2, √(3)/2)

2. (a) (i) 8π/3 (ii) 1 (iii) II (iv) (-1/2, √(3)/2)

(b) (i) 11π/3 (ii) 1 (iii) IV (iv) (1/2, -√(3)/2)

(c) (i) -17π/3 (ii) 2 (iii) I (iv) (-1/2, √(3)/2)

3. (i) 510°

(ii) 1

(iii) II

(iv) (-√(3)/2, 1/2)

4. (a) (i) 1350° (ii) 3 (iii) on negative y-axis (iv) (0, -1)

(b) (i) -2250° (ii) 6 (iii) IV (iv) (√(2)/2, -√(3)/2)

(c) (i) 1590° (ii) 4 (iii) II (iv) (-√(3)/2, 1/2)

5. (a) 1 (b) 1 (c) 1 (d) 1

6. ???

7. (a) yes, because (3/5)

Here are the answers to your homework assignment:

Working with the Unit Circle Exercises - Answers

1. (i) 7π/3

(ii) 1

(iii) I

(iv) (1/2, √(3)/2)

2. (a) (i) 8π/3 (ii) 1 (iii) II (iv) (-1/2, √(3)/2)

(b) (i) 11π/3 (ii) 1 (iii) IV (iv) (1/2, -√(3)/2)

(c) (i) -17π/3 (ii) 2 (iii) I (iv) (-1/2, √(3)/2)

3. (i) 510°

(ii) 1

(iii) II

(iv) (-√(3)/2, 1/2)

4. (a) (i) 1350° (ii) 3 (iii) on negative y-axis (iv) (0, -1)

(b) (i) -2250° (ii) 6 (iii) IV (iv) (√(2)/2, -√(3)/2)

(c) (i) 1590° (ii) 4 (iii) II (iv) (-√(3)/2, 1/2)

5. (a) 1 (b) 1 (c) 1 (d) 1

6. ???

7. (a) yes, because (3/5)

^{2}+ (4/5)^{2}= 1 (b) no (c) yes(d) yes
Labels:
Circular Functions,
Mr. Kuropatwa,
Slides

### The Scribe List

This is

This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.

3.23

**The Scribe List**. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.

IMPORTANT: Make sure you label all your Scribe Posts properly or they will not be counted.

__Cycle 3__Francis Joyce Eleven benofschool roxanne | JamieNeRd123C zeph Richard | nelsa Rence kristina Paul |

3.23

## Thursday, February 14, 2008

Hello everyone I am Richard and I am your scribe for Valentines day

Today we learned that there are four different ways to write a function but they all mean exactly the same thing, but they are just 4 different points of view. All the examples have to do with Mr. K's pet otter and that he eats two fish a day

1) In words

Example My pet Otter eats two fish a day.

2) In A table , you would draw it out in a table

Example below

3) in a Graph, You would graph out the information

example below

4) In an Equation you would make an equation for your data.

Example : y = 2x

Slide two

we did some problems

jsut a reminder dont when solving for the cos or sin or tan of a number and you find it you dont have to put the sin, cos ,tan in front of it anymore.

example cos² ( 5π/6) → cos² (-√3/2 ) which is wrong

the correct way is cos² ( 5π/6) → (-√3/2 )

After we did ten mental math questions

on slide 6 we also did some questions we didn't have enough time to do all of them

and the questions that you did not do is for homework.

☺☺☺☺☺☺☺☺☺☺☺

In the afternoon Mr K was away because he had to go to a confrence to sspeak about leadership. the sub's name was Ms. Hallson. she gave us an assignment and that it was due tommorow.

well that was my scribe i hope it was usefull.... the nest scribe is going to be roxanne

Today we learned that there are four different ways to write a function but they all mean exactly the same thing, but they are just 4 different points of view. All the examples have to do with Mr. K's pet otter and that he eats two fish a day

1) In words

Example My pet Otter eats two fish a day.

2) In A table , you would draw it out in a table

Example below

3) in a Graph, You would graph out the information

example below

4) In an Equation you would make an equation for your data.

Example : y = 2x

Slide two

we did some problems

jsut a reminder dont when solving for the cos or sin or tan of a number and you find it you dont have to put the sin, cos ,tan in front of it anymore.

example cos² ( 5π/6) → cos² (-√3/2 ) which is wrong

the correct way is cos² ( 5π/6) → (-√3/2 )

After we did ten mental math questions

on slide 6 we also did some questions we didn't have enough time to do all of them

and the questions that you did not do is for homework.

☺☺☺☺☺☺☺☺☺☺☺

In the afternoon Mr K was away because he had to go to a confrence to sspeak about leadership. the sub's name was Ms. Hallson. she gave us an assignment and that it was due tommorow.

well that was my scribe i hope it was usefull.... the nest scribe is going to be roxanne

Labels:
Circular Functions,
richard,
Scribe

## Wednesday, February 13, 2008

### Circular Functions (February 13, 2008)

Well guys it's me, Justus here, and as the scribe for today I'm writing this blog post :)

To begin I'd like to say that we had a substitute today, a very nice lady by the name of, Ms. Cheekie. Anyways after introducing herself, she informed us that we had a quiz on circular functions, which we had the whole class to do, but must be handed in at the end of the class. If you missed the class I'd sugges talking to Mr. Kuropatwa and seeing if you'd be able to write the quiz, as I assume its for marks.

After that, we were handed out a worksheet on circular functions, entitled, "Working With the Unit Circle Exercises." Which was to be worked on once completed the quiz. If we didnt finish the worksheet during regular class time, Ms. Cheekie said that it would be due during TOMORROWS CLASS (as in February 14th, 2008, at 9:00am.)

That basically sums up what happened during todays class, and thus, sums up my scribe post. However, because this scribe was significantly shorter then most posts, and I'm not entirely sure if this material counted as a legitimate scribe post (regardless of the fact that we do need to know of the hand in sheet), I decided to just post anyways.

That is all, until next time.

To begin I'd like to say that we had a substitute today, a very nice lady by the name of, Ms. Cheekie. Anyways after introducing herself, she informed us that we had a quiz on circular functions, which we had the whole class to do, but must be handed in at the end of the class. If you missed the class I'd sugges talking to Mr. Kuropatwa and seeing if you'd be able to write the quiz, as I assume its for marks.

After that, we were handed out a worksheet on circular functions, entitled, "Working With the Unit Circle Exercises." Which was to be worked on once completed the quiz. If we didnt finish the worksheet during regular class time, Ms. Cheekie said that it would be due during TOMORROWS CLASS (as in February 14th, 2008, at 9:00am.)

That basically sums up what happened during todays class, and thus, sums up my scribe post. However, because this scribe was significantly shorter then most posts, and I'm not entirely sure if this material counted as a legitimate scribe post (regardless of the fact that we do need to know of the hand in sheet), I decided to just post anyways.

**Should it be unsatisfactory**as a scribe post, I will help Richard (whom I've selected as the next scribe) out with tomorrows scribe (since it is a two period day.)That is all, until next time.

Labels:
Circular Functions,
Justus,
Scribe

## Tuesday, February 12, 2008

### SOLVING TRIGONOMETRIC EQUATIONS

Hello world. I am Jamie and I come in peace. I’m stranded in a lodge somewhere, looking out through a window, perhaps the one that you see on this blog page and I’m feeling rather…snog. Haha. The wonders of the English language— everyday there is a word of the day in the room. Yesterday was picayune and today’s was petard. I was just wondering if he was doing this in descending alphabetical order. Sometimes I just go into TANGENTS and look for random things around the room and there I see…a word…for the day…it just magically changes.

Anyhow, in all seriousness [kind of], today’s first SECTOR of math notes was all about SOLVING TRIGONOMETRIC EQUATIONS. Half of the class time was spent reviewing content that was from the grade 11 curriculum in regards to finding the smallest angle formed by the terminal arm [the arm which extends from the origin point of a Cartesian plane to a point on the grid] and made with the x-axis. This is known as the RELATED ANGLE.

The class was then instructed to plug in examples to find the value of a certain angle, usually represented by “θ”. An example was 2 sin θ = 1. The first thing to do is similar with dealing with a variable. In order to find the value of a variable, it needs to be isolated, therefore sin θ = ½. One half is positive so the sine value has to be positive. Knowing this allows one to know which two quadrants share this positive sine value; in quadrants I and II. I’m sorry to say Mr. K, determining the quadrants by CAST is still relevant to me, and we gotta love it. It is after all Valentine’s Day [soon, at least.]. Like all technology, even a calculator has glitches. Determining the quadrant also indicates where the terminal arm is and when plugging in the value in the calculator, the calculator only gives one answer which is the related angle, but this answer is also the REFERENCE ANGLE [θref]. It is called the reference angle since it refers to either adding or subtracting this value from the value on the x or y-axis, depending on the rules of how to find the angle’s value in the quadrant. In quadrant I, θ = ½ at π/6. To find the angle in quadrant II, you would have to subtract π/6 from π which happens to give you 5 π/6. θ = π/6 and 5π/6.

The class did many problems similar to that, but before advancing, we habitual humans had to do our third mental math exercise on our unit circles. THINK IN RADIANS everyone!! To celebrate, just do the clock dance.

Immediately after the refresher course, the easy related angle concept slowly adapted into a more complex one. Again, everyone had to look back and dwell on their factoring skills. But in alternative of factoring variables and isolating x in binomials and trinomials, we instead isolate sine, cosine or tangent of x.

ex. Solve for x on the interval [0, 2π] in 2 sin2x = sin x. The first thing to do is transpose sin x to the other side so that one side equals zero. Then factor sin x out of the equation to simplify, making it easier to find the sine value of x. Finally, find the measure of the angles using these values and in RADIANS!!

2 sin2x = sin x à 2 sin2x - sin x = 0 à sin x (2 sin x – 1) à sin x = 0, ½

Therefore x = 0, π, 2π and π/6 and 5π/6.

**NOTE if you cannot see how to factor this, you can also substitute sin x with another value like “a” [or Helena Bonham Carter, Queen Elizabeth, Johnny Depp, Jamie, etc…whatever floats your luxury yacht.] just as long as it is not the same variable like x because x does not EQUAL sin x. sin x is a function of x, therefore a manipulated x cannot equal x. Just remember to substitute back what you substituted in the beginning to get full marks and clarify what values are.

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

In the afternoon, we continued going over the same material in order to receive a thorough understanding on the idea. As usual though, we took it a step further.

Do you ever open up a calculus book and wonder what those symbols mean? I do, but I’m just the big nerd. Hence, Jamie123C was born. That’s what you do when you don’t have wonders like…I don’t know…the internet around you. Flabberjacks, I still don’t have internet. How am I blogging then, you say? It’s all magic. I’m not lying. I went to Hogwarts and everything. Everything was swell there, I had an owl, people were chipper waving around their magic wands. Mine was the root of a phoenix feather. Not kidding. Here I am again— my mind going astray off to Fleet Street or some place like that. Oh, Sweeney Todd.

Anyhow, I think I’m finished, I’m no longer derailed on my train of thought. Where was I? Ah, yes. Symbols— they just look like the DaVinci code or something. As each of us grow mathematically, we come across the meaning of some of them. Starting off with “x Ɛ R”. This is familiar in terms of finding the domain. So instead of finding the values of sine of x in between [0, 2π], which happens to be the distance of revolution around the circle once, the domain will be x Ɛ R which circulates the circle multiple times, infinite times because it has no restriction in domain. But having limited time which should be spent like gold, we can’t count all of the values which apply. So we have to find a pattern and make an equation just as shown in the following example:

cos2 x – 2 cos x = 0 à cos x (cos x – 2) = 0 à cos x = 0 and 2.

***cos x = 2 is EXTRANEOUS because the cosine of 2 is undefined since cosine value cannot be more than one. The same thing applies for sine. Tangent on the other hand can be more than one. [sinθ <> 1] Further explanation of this is on the toy, patent pending— erm, math tool: Dave’s Unit Circle Applet, link is found in slides for today.

Therefore x = π/6 and 5π/6

But since there is no restriction on the domain, possible values could be 13π/6, etc…

x = π/6 + 2kπ; k Ɛ I OR in this situation, x = 5π/6 + 2kπ; k Ɛ I.

where k represents a variable that is always a whole number and an integer to find out how many times you can go around the unit circle and find the same value and k is an element of an integer…

This scribe post is a little lengthy and droning, especially with the jokes and lessons merged together. You know, I try. I think I’ve summed up today. My bloggerish mood has begun to fade, so who shall take on these powers tomorrow? The answer to that is in this question…[sort of an inside joke to the math room] Let’s just pretend I’m situated in quadrant two and there are three other people around me. In which quadrant is both tanθ > 0 and sinθ < 0? ANSWER? QUADRANT III!! So that means Justus is next. Sorry man, it’s just that you’re adjacent to me and it worked, math-wise… I try to make every experience educational. Hey, who knows, it could’ve been Queen Elizabeth sitting behind me and scribing tomorrow.

oh yeah. pps I adore you [sequel to the ps I love you, going to be a greater movie] I BElieve homework tonight is Exercise 5

Anyhow, in all seriousness [kind of], today’s first SECTOR of math notes was all about SOLVING TRIGONOMETRIC EQUATIONS. Half of the class time was spent reviewing content that was from the grade 11 curriculum in regards to finding the smallest angle formed by the terminal arm [the arm which extends from the origin point of a Cartesian plane to a point on the grid] and made with the x-axis. This is known as the RELATED ANGLE.

The class was then instructed to plug in examples to find the value of a certain angle, usually represented by “θ”. An example was 2 sin θ = 1. The first thing to do is similar with dealing with a variable. In order to find the value of a variable, it needs to be isolated, therefore sin θ = ½. One half is positive so the sine value has to be positive. Knowing this allows one to know which two quadrants share this positive sine value; in quadrants I and II. I’m sorry to say Mr. K, determining the quadrants by CAST is still relevant to me, and we gotta love it. It is after all Valentine’s Day [soon, at least.]. Like all technology, even a calculator has glitches. Determining the quadrant also indicates where the terminal arm is and when plugging in the value in the calculator, the calculator only gives one answer which is the related angle, but this answer is also the REFERENCE ANGLE [θref]. It is called the reference angle since it refers to either adding or subtracting this value from the value on the x or y-axis, depending on the rules of how to find the angle’s value in the quadrant. In quadrant I, θ = ½ at π/6. To find the angle in quadrant II, you would have to subtract π/6 from π which happens to give you 5 π/6. θ = π/6 and 5π/6.

The class did many problems similar to that, but before advancing, we habitual humans had to do our third mental math exercise on our unit circles. THINK IN RADIANS everyone!! To celebrate, just do the clock dance.

Immediately after the refresher course, the easy related angle concept slowly adapted into a more complex one. Again, everyone had to look back and dwell on their factoring skills. But in alternative of factoring variables and isolating x in binomials and trinomials, we instead isolate sine, cosine or tangent of x.

ex. Solve for x on the interval [0, 2π] in 2 sin2x = sin x. The first thing to do is transpose sin x to the other side so that one side equals zero. Then factor sin x out of the equation to simplify, making it easier to find the sine value of x. Finally, find the measure of the angles using these values and in RADIANS!!

2 sin2x = sin x à 2 sin2x - sin x = 0 à sin x (2 sin x – 1) à sin x = 0, ½

Therefore x = 0, π, 2π and π/6 and 5π/6.

**NOTE if you cannot see how to factor this, you can also substitute sin x with another value like “a” [or Helena Bonham Carter, Queen Elizabeth, Johnny Depp, Jamie, etc…whatever floats your luxury yacht.] just as long as it is not the same variable like x because x does not EQUAL sin x. sin x is a function of x, therefore a manipulated x cannot equal x. Just remember to substitute back what you substituted in the beginning to get full marks and clarify what values are.

~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~

In the afternoon, we continued going over the same material in order to receive a thorough understanding on the idea. As usual though, we took it a step further.

Do you ever open up a calculus book and wonder what those symbols mean? I do, but I’m just the big nerd. Hence, Jamie123C was born. That’s what you do when you don’t have wonders like…I don’t know…the internet around you. Flabberjacks, I still don’t have internet. How am I blogging then, you say? It’s all magic. I’m not lying. I went to Hogwarts and everything. Everything was swell there, I had an owl, people were chipper waving around their magic wands. Mine was the root of a phoenix feather. Not kidding. Here I am again— my mind going astray off to Fleet Street or some place like that. Oh, Sweeney Todd.

Anyhow, I think I’m finished, I’m no longer derailed on my train of thought. Where was I? Ah, yes. Symbols— they just look like the DaVinci code or something. As each of us grow mathematically, we come across the meaning of some of them. Starting off with “x Ɛ R”. This is familiar in terms of finding the domain. So instead of finding the values of sine of x in between [0, 2π], which happens to be the distance of revolution around the circle once, the domain will be x Ɛ R which circulates the circle multiple times, infinite times because it has no restriction in domain. But having limited time which should be spent like gold, we can’t count all of the values which apply. So we have to find a pattern and make an equation just as shown in the following example:

cos2 x – 2 cos x = 0 à cos x (cos x – 2) = 0 à cos x = 0 and 2.

***cos x = 2 is EXTRANEOUS because the cosine of 2 is undefined since cosine value cannot be more than one. The same thing applies for sine. Tangent on the other hand can be more than one. [sinθ <> 1] Further explanation of this is on the toy, patent pending— erm, math tool: Dave’s Unit Circle Applet, link is found in slides for today.

Therefore x = π/6 and 5π/6

But since there is no restriction on the domain, possible values could be 13π/6, etc…

x = π/6 + 2kπ; k Ɛ I OR in this situation, x = 5π/6 + 2kπ; k Ɛ I.

where k represents a variable that is always a whole number and an integer to find out how many times you can go around the unit circle and find the same value and k is an element of an integer…

This scribe post is a little lengthy and droning, especially with the jokes and lessons merged together. You know, I try. I think I’ve summed up today. My bloggerish mood has begun to fade, so who shall take on these powers tomorrow? The answer to that is in this question…[sort of an inside joke to the math room] Let’s just pretend I’m situated in quadrant two and there are three other people around me. In which quadrant is both tanθ > 0 and sinθ < 0? ANSWER? QUADRANT III!! So that means Justus is next. Sorry man, it’s just that you’re adjacent to me and it worked, math-wise… I try to make every experience educational. Hey, who knows, it could’ve been Queen Elizabeth sitting behind me and scribing tomorrow.

oh yeah. pps I adore you [sequel to the ps I love you, going to be a greater movie] I BElieve homework tonight is Exercise 5

Labels:
Circular Functions,
Jamie,
Scribe

## Monday, February 11, 2008

### Trigonometric Equations

Alrightey I finally got home. Sorry for the late post had some technical difficulties with the computer also. Anyways here's the recap of what we did on todays slides.

Slide 2:

Here we see the reciprocal trig functions we learned about yesterday. Just for review, the new trig functions we learned were cosecant, secant, and cotangent. Mr. K emphasized that most kids go wrong when they think that cosecant is with cos and secant is with sin. So make sure to not do that! Anyways as it says on the slide, we get cosecant by 1/sin(theta), secant by 1/cos(theta), and cotangent by 1/tan(theta). Or we can just change the numerator with the denominator to save time. As I learned today, don't switch any signs. If it's negative in the denominator it will still be negative when it becomes your numerator when you flip the fraction.

Slide 3:

Ok the question here was a bit of a review from the questions we've been doing for the past few days. First we must find the hypotenuse of the big triangle. We do that by plotting the point (9,-40), which are your x and y values. Then we must find the cos and sin values. We do this by finding the angles adjacent to theta over the hypotenuse (for the cos), and for sin we put the angle opposite of theta over the hypotenuse. The cos and sin value you get will be the coordinates for theta! As for the second question on the slide, just list the cos, sin, tan, secant, cosecant, and cotangent values. Remember for the reciprocal trig functions, just flip the fraction.

Slide 5:

We've been practicing how to find the exact values with questions like these on this slide. Doing these questions are a lot easier if you know the values around the unit circle. K i'll make this brief, all you have to do is substitue the exact values in and simplify!

Slide 6:

Oh yes I remember this slide from this morning. Mr. K challenged our mathematical minds to do these questions. But like he said there's a method to his madness or something like that. I think we all know how to do these kinds of questions. We just isolate x or factor.

Slide 7:

We now see what Mr. K was trying to do by starting us off easy. We do exactly what we did on the last slide. We simply isolate sinx and get the value of 1/2. Now what on the unit circle do we know has a y value of 1/2? pi/6, and 5pi/6 of course.

Slide 8:

1+2cosx = 5cosx

The question gets a bit tougher but we still do the same things. Isolate cosx and it will equal 1/3. Now we take our trusty calculators and press 2nd fnc cos then 1/3. Make sure to be on radians. We then get the value 1.2309. But there is another value because cos is also positive in quadrant 4. How do we get the value in quadrant 4? Just put in your calculator 2pi - 1.2309 to get the other value of x, which is 5.0522 the related angle. K a quick summary, if looking for related angle in q2 you take the reference angle and subtract it from pi. If looking for related angle in q3 you take the reference angle and add it to pi. And that's where we ended!

Hope you guys understood my rambling. Oh yeah next scriiibe is...........................JAMIE since I took your turn today. KK goodniight everyone *faints*.

Slide 2:

Here we see the reciprocal trig functions we learned about yesterday. Just for review, the new trig functions we learned were cosecant, secant, and cotangent. Mr. K emphasized that most kids go wrong when they think that cosecant is with cos and secant is with sin. So make sure to not do that! Anyways as it says on the slide, we get cosecant by 1/sin(theta), secant by 1/cos(theta), and cotangent by 1/tan(theta). Or we can just change the numerator with the denominator to save time. As I learned today, don't switch any signs. If it's negative in the denominator it will still be negative when it becomes your numerator when you flip the fraction.

Slide 3:

Ok the question here was a bit of a review from the questions we've been doing for the past few days. First we must find the hypotenuse of the big triangle. We do that by plotting the point (9,-40), which are your x and y values. Then we must find the cos and sin values. We do this by finding the angles adjacent to theta over the hypotenuse (for the cos), and for sin we put the angle opposite of theta over the hypotenuse. The cos and sin value you get will be the coordinates for theta! As for the second question on the slide, just list the cos, sin, tan, secant, cosecant, and cotangent values. Remember for the reciprocal trig functions, just flip the fraction.

Slide 5:

We've been practicing how to find the exact values with questions like these on this slide. Doing these questions are a lot easier if you know the values around the unit circle. K i'll make this brief, all you have to do is substitue the exact values in and simplify!

Slide 6:

Oh yes I remember this slide from this morning. Mr. K challenged our mathematical minds to do these questions. But like he said there's a method to his madness or something like that. I think we all know how to do these kinds of questions. We just isolate x or factor.

Slide 7:

We now see what Mr. K was trying to do by starting us off easy. We do exactly what we did on the last slide. We simply isolate sinx and get the value of 1/2. Now what on the unit circle do we know has a y value of 1/2? pi/6, and 5pi/6 of course.

Slide 8:

1+2cosx = 5cosx

The question gets a bit tougher but we still do the same things. Isolate cosx and it will equal 1/3. Now we take our trusty calculators and press 2nd fnc cos then 1/3. Make sure to be on radians. We then get the value 1.2309. But there is another value because cos is also positive in quadrant 4. How do we get the value in quadrant 4? Just put in your calculator 2pi - 1.2309 to get the other value of x, which is 5.0522 the related angle. K a quick summary, if looking for related angle in q2 you take the reference angle and subtract it from pi. If looking for related angle in q3 you take the reference angle and add it to pi. And that's where we ended!

Hope you guys understood my rambling. Oh yeah next scriiibe is...........................JAMIE since I took your turn today. KK goodniight everyone *faints*.

Labels:
Circular Functions,
Joyce,
Scribe

## Sunday, February 10, 2008

### The Great Almighty Unit Circle. Plus other stuff .. That can be almighty if you want it to. ^^

Hey everybody! Sorry for the two days and a half late post. Anyways, what I'm going to be discussing in my blog is what we've done that day. First, we started it off by finding that there is a super Pi on the Chinese coin. After that we discussed the points on the Unit Circle. What was Sinθ, Cosθ, and Tanθ were on each point of the Unit Circle. Then we discovered patterns and methods of remembering the whole unit circle.

The unit circle is called the unit circle because the radius is 1. The angles we used in the unit circle was 30°, 45° and 60°. We converted degrees into radians. 30° = Pi/6, 45° = Pi/4 and 60° = Pi/3.

**

- In a right angle triangle, we know that Sinθ = OPP / HYP, Cosθ = ADJ / HYP, and Tanθ = OPP / ADJ. SOH CAH TOA

- We then find that Tanθ = Sinθ / Cosθ

** In the unit circle, Pi/6, if you drop straight down from P(θ) to form a right angle triangle, you can find the coordinates of P(θ). As you can see in my professional diagram below.

The coordinates of P(θ) is (Cosθ, Sinθ) .. Because we know the unit circle's radius is 1, and the hypotenuse is 1, then we can figure out Sinθ, and Cosθ, using Pi/6 (30 Degrees) ..

- Sin Pi/6 = 1/2 - Cos Pi/6 = v3/2

** Keep in mind that it will always be over 2 So P(θ) = P(v3/2, 1/2) Knowing that these are the points that represent P(θ) then that must mean that Cosθ is represented by the X axis and Sinθ is represented by the Y axis. -Tanθ = y/x

** That is how we found the points for the unit circle.

- A short lesson we learned was complementary angles.

- It is the sum of 2 angles that add up to 90°

Ex. 30° + 60° = 90°

Sin30° = 1/2

Cos60° = 1/2

** The unit circle can be found on the last lesson's slide (The one with Happy New Year!!). Slide #6/8 .. It is a very good unit circle.

-The method of remembering each point is you have to remember 3 numbers. 1, 2, and 3.

Here is a link to my amazing, incredibly, outstanding paint skills for a clear example of one way of remembering the unit circle.

Click Here!

The explaining will be here.

-The black numbers represent Sinθ

-The red numbers represent Cosθ

-The green numbers represent Tanθ

What do you notice?

-RIGHT! 1, 2 and 3. (You're probably thinking in your head .. "uh.. no")

**Anyways

The numbers you see will always be square rooted over 2.

Yes, Square root of 3, over 2

-Square root of 2, over 2

-Square root of 1, over 2.

**Now how do I know when the number is 1, 2 or 3?

The purple lines in the circle represent the Y axis, also Sinθ.

Ditto for the blue lines, it represents the X axis, Cosθ.

**What do you notice?

-RIGHT! That Pi/4 or 45° will always be square root of 2 over 2. (Not including Tangent)

As you see the longer the lines are extended, it is square root of 3 over 2.

The shorter it is, it is 1 half.

Between that is square root of 2 over 2.

Yea the clock dance becomes useful. Unfortunately.

**Now that we have figured out Sinθ and Cosθ .. How about Tanθ?

Well .. You should have already picked up that the middle numbers are always 1.

**The closer tangent is to the X axis, it is under 1. Yes 1 over square root of 3.

The farther it is, (over the X axis) it is square root of 3 over 1. Or just square root of 3.

**I hope this helped. I was in a rush.

I'll skip to to the credits.

**The next scribe will be the person who sits the closest to me.

Yup, you got it right again! Joyce will be the next scribe =)

This is Agent Eleven, out.

Better days.

The unit circle is called the unit circle because the radius is 1. The angles we used in the unit circle was 30°, 45° and 60°. We converted degrees into radians. 30° = Pi/6, 45° = Pi/4 and 60° = Pi/3.

**

- In a right angle triangle, we know that Sinθ = OPP / HYP, Cosθ = ADJ / HYP, and Tanθ = OPP / ADJ. SOH CAH TOA

- We then find that Tanθ = Sinθ / Cosθ

** In the unit circle, Pi/6, if you drop straight down from P(θ) to form a right angle triangle, you can find the coordinates of P(θ). As you can see in my professional diagram below.

The coordinates of P(θ) is (Cosθ, Sinθ) .. Because we know the unit circle's radius is 1, and the hypotenuse is 1, then we can figure out Sinθ, and Cosθ, using Pi/6 (30 Degrees) ..

- Sin Pi/6 = 1/2 - Cos Pi/6 = v3/2

** Keep in mind that it will always be over 2 So P(θ) = P(v3/2, 1/2) Knowing that these are the points that represent P(θ) then that must mean that Cosθ is represented by the X axis and Sinθ is represented by the Y axis. -Tanθ = y/x

** That is how we found the points for the unit circle.

- A short lesson we learned was complementary angles.

- It is the sum of 2 angles that add up to 90°

Ex. 30° + 60° = 90°

Sin30° = 1/2

Cos60° = 1/2

** The unit circle can be found on the last lesson's slide (The one with Happy New Year!!). Slide #6/8 .. It is a very good unit circle.

-The method of remembering each point is you have to remember 3 numbers. 1, 2, and 3.

Here is a link to my amazing, incredibly, outstanding paint skills for a clear example of one way of remembering the unit circle.

Click Here!

The explaining will be here.

-The black numbers represent Sinθ

-The red numbers represent Cosθ

-The green numbers represent Tanθ

What do you notice?

-RIGHT! 1, 2 and 3. (You're probably thinking in your head .. "uh.. no")

**Anyways

The numbers you see will always be square rooted over 2.

Yes, Square root of 3, over 2

-Square root of 2, over 2

-Square root of 1, over 2.

**Now how do I know when the number is 1, 2 or 3?

The purple lines in the circle represent the Y axis, also Sinθ.

Ditto for the blue lines, it represents the X axis, Cosθ.

**What do you notice?

-RIGHT! That Pi/4 or 45° will always be square root of 2 over 2. (Not including Tangent)

As you see the longer the lines are extended, it is square root of 3 over 2.

The shorter it is, it is 1 half.

Between that is square root of 2 over 2.

Yea the clock dance becomes useful. Unfortunately.

**Now that we have figured out Sinθ and Cosθ .. How about Tanθ?

Well .. You should have already picked up that the middle numbers are always 1.

**The closer tangent is to the X axis, it is under 1. Yes 1 over square root of 3.

The farther it is, (over the X axis) it is square root of 3 over 1. Or just square root of 3.

**I hope this helped. I was in a rush.

I'll skip to to the credits.

**The next scribe will be the person who sits the closest to me.

Yup, you got it right again! Joyce will be the next scribe =)

This is Agent Eleven, out.

Better days.

Labels:
Circular Functions,
Eleven,
Scribe

## Friday, February 8, 2008

## Thursday, February 7, 2008

### Exact Values In The Unit Circle

Hello everyone, I'm Nelsa, today's scribe. I'm so sorry for the late post, Thursday is a really busy day for me.

Mr. Kuropatwa began the class by asking everyone if we have read the 'Digital Ethics' post and taking everyone's oath (by shaking our hands) that we'll abide by those rules. He also explained 'Blogging On Blogging' - which you can learn more about by reading that post - and how to properly label your posts. But of course, this is a math class, and that's probably what you all are waiting for. Most of the things we studied today, reviewed a lot of what we learned yesterday, so I'm sorry if it seems like I'm repeating the things Ben has detailed on the last scribe post.

The first slide required us to find which quadrant P(5) can be found in. Most of us were more than slightly confused because of the lack of a y-coordinate, but Mr. K simply told us to think in 'radians'. With this in mind, we reviewed what we already knew, which was, 180°=π and 360°=2π. π is approximately 3, which means that 2π would be approximately 6. In order to get 5, we subtract one from six, and whenever we subtract, we move in a clockwise direction, which would mean that P(5) is in quadrant IV.

We were also asked to find which quadrant contained sinθ <> 0. To do this, Mr. K told us to forget all about 'CAST', which I'm sure we all learned last year, and instead, understand what's going on. Cosine, as we know from 'SOHCAHTOA', is the x-axis or coordinate, and sine, is the y-axis or coordinate. Keeping this in mind, we know that in quadrant I, both cosine and sine is positive, as both the x-axis and the y-axis is positive. This also means that tangent is positive, because when you divide two positives (which is really what tangent is, sine/cosine), it's a positive. In the second quadrant, cosine is negative, because it's before the zero, but sine is still positive, which means that tangent is negative. In the third quadrant, both cosine and sine is negative, as both are before the zero, which means tangent is positive. Finally, in the fourth quadrant, cosine is positive, and sine is negative, making tangent negative as well. Using this information, we were able to answer the first question on the second slide.

The equation of a unit circle is x(squared) + y(squared) = 1. The third slide asked us to determine whether the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. So using the above formula, we figured out that the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. But a simpler way to find the answer is to understand, again, that cosine is the x-coordinate, and sine is the y-coordinate. The radius of a unit circle is 1 (hence 'unit'), so the cosine squared, plus the sine squared should equal to 1, which is the hypotenuse.

Mr. K then went on to talk about the person, who died because of one of the triangles included in your basic geometry set. I forgot the person's name, but basically, a group of people who call themselves the Pythagoreans believed that the world was made up of (only) rational numbers. One of the triangles in a geometry set, besides having a 90° angle, also have 45° angles. The hypotenuse of this triangle, is the root of 2, which is an in irrational number. The Pythagoreans were aghast, and kept this fact hidden from the rest of the world, as it would ruin everything that they believed in. But one person from the inner circle (who were the only people who knew) told everyone else, and, to make a long story short, was thrown off a cliff.

So that's basically what we learned in class today. Again, I'm so, so sorry for the late post. The next scribe is twenty-seven hundred, ninety-eight, twenty, ELEVEN, nine. =)

Mr. Kuropatwa began the class by asking everyone if we have read the 'Digital Ethics' post and taking everyone's oath (by shaking our hands) that we'll abide by those rules. He also explained 'Blogging On Blogging' - which you can learn more about by reading that post - and how to properly label your posts. But of course, this is a math class, and that's probably what you all are waiting for. Most of the things we studied today, reviewed a lot of what we learned yesterday, so I'm sorry if it seems like I'm repeating the things Ben has detailed on the last scribe post.

The first slide required us to find which quadrant P(5) can be found in. Most of us were more than slightly confused because of the lack of a y-coordinate, but Mr. K simply told us to think in 'radians'. With this in mind, we reviewed what we already knew, which was, 180°=π and 360°=2π. π is approximately 3, which means that 2π would be approximately 6. In order to get 5, we subtract one from six, and whenever we subtract, we move in a clockwise direction, which would mean that P(5) is in quadrant IV.

We were also asked to find which quadrant contained sinθ <> 0. To do this, Mr. K told us to forget all about 'CAST', which I'm sure we all learned last year, and instead, understand what's going on. Cosine, as we know from 'SOHCAHTOA', is the x-axis or coordinate, and sine, is the y-axis or coordinate. Keeping this in mind, we know that in quadrant I, both cosine and sine is positive, as both the x-axis and the y-axis is positive. This also means that tangent is positive, because when you divide two positives (which is really what tangent is, sine/cosine), it's a positive. In the second quadrant, cosine is negative, because it's before the zero, but sine is still positive, which means that tangent is negative. In the third quadrant, both cosine and sine is negative, as both are before the zero, which means tangent is positive. Finally, in the fourth quadrant, cosine is positive, and sine is negative, making tangent negative as well. Using this information, we were able to answer the first question on the second slide.

The equation of a unit circle is x(squared) + y(squared) = 1. The third slide asked us to determine whether the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. So using the above formula, we figured out that the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. But a simpler way to find the answer is to understand, again, that cosine is the x-coordinate, and sine is the y-coordinate. The radius of a unit circle is 1 (hence 'unit'), so the cosine squared, plus the sine squared should equal to 1, which is the hypotenuse.

Mr. K then went on to talk about the person, who died because of one of the triangles included in your basic geometry set. I forgot the person's name, but basically, a group of people who call themselves the Pythagoreans believed that the world was made up of (only) rational numbers. One of the triangles in a geometry set, besides having a 90° angle, also have 45° angles. The hypotenuse of this triangle, is the root of 2, which is an in irrational number. The Pythagoreans were aghast, and kept this fact hidden from the rest of the world, as it would ruin everything that they believed in. But one person from the inner circle (who were the only people who knew) told everyone else, and, to make a long story short, was thrown off a cliff.

So that's basically what we learned in class today. Again, I'm so, so sorry for the late post. The next scribe is twenty-seven hundred, ninety-eight, twenty, ELEVEN, nine. =)

### Blogging On Blogging (BOB)

We were talking about exactly what sort of post you're supposed to make to get that one blogging mark on your test. The kind of post I'd like you to make should have one or more of these characteristics:

Your posts do not have to be long. I'm far more interested in the

Make certain you always use 3 labels on your post:

When you share where you are in your learning a few days before the unit test I can address those issues in class so, hopefully, you will get much more than one extra mark on the test. ;-)

- Write about what you understand the least in the unit so far; your personal "Muddiest Point."
- A reflection on a particular class.
- A reflective comment on your progress in the course.
- A comment on something that you've learned that you thought was "cool".
- A comment about something that you found very hard to understand but now you get it! Describe what sparked that "moment of clarity" and what it felt like.
- Have you come across something we discussed in class out there in the "real world" or another class? Describe the connection you made.

Your posts do not have to be long. I'm far more interested in the

**quality**of what you write rather than the**quantity**.Make certain you always use 3 labels on your post:

**[your name], [unit tag], BOB**When you share where you are in your learning a few days before the unit test I can address those issues in class so, hopefully, you will get much more than one extra mark on the test. ;-)

*Happy Blogging!*## Wednesday, February 6, 2008

### Circular Funtions and Circles On a Cartesian Plain

Okay I hello everybody, this is benofschool with today's scribe on what we did. We started with talking about Digital Ethics or things to be careful of on the internet. Watch the video on the blog to find out more.

Mr.K also told us what new things to look for on the blog in the near future. Such features include a walkie-talkie sort of thing found on the blog which allows people with a microphone on the computer to send voice messages to other people who are online on the blog. Another cool feature is a little number in the corner of the blog which shows the number of people online which helps with using the walkie-talkie feature. The last but not least feature is a chat/shout box which allows us to chat. But with these features also come trust. If we use these features we must follow the Digital Ethics Rules and keep things appropriate. Remember don't post anything that you don't want anybody to know.

Okay onto the math. We did a small review on what we did yesterday. One reminder about units. There are no units for radian answers. All of the units used in finding the radian answer have been reduced to 1 which doesn't change any answer so it is as if they are not there. One more thing, the equation to convert degrees to radians and vice versa is not a formula but a proportion. It shows how degree values are proportional to radian values. So don't memorize that exact equation but remember how degree values are proportional to radian values.

The first question we worked on is on the second slide on the Feb.6 slide post. One solution is showed on the slide but there was one trick that could be used. We know that a full circle is 2π and we can find out that how the hands are found on a clock at 4:00 that it is one-third of the clock or the full circle. So just divide 2π by 3 which gives us our answer with little work.

The second question we worked on is found on the third slide. (Ignore that little message at the bottom at the slide it is irrelevant). That is the correct solution to the problem and I will now explain the process. To answer the problem you must find the area of the circle using the area of a circle equation. Then make a proportional equation like the one on the slide. It shows that 30° out of 360° is equal to the unknown area out of the total area of the circle. Now all that you have to do is solve for the unknown or on the slide, A. Remember to include units where necessary.

That ended our first class for today. Notice that I said first. There was another one.

In our second class we continued with a new problem and the first question required us to know what the word "coterminal" means. Coterminal means end together. So in the question we need to find out what other radian ends where -π/3 (I don't know how to write fractions on the blog) ends or the positive coterminal of -π/3. So we know that a full circle in radians is 2π. we need to subtract π/3 to 2π to find where it ends. If you are wondering what a negative angle is, a negative angle is the same as going backwards from 360° or 2π. One example would be -50°. That is 360°-50°= 310°. Same goes with radians in this problem. We need to subtract π/3 from 2π. By using basic fraction subtracting skills we can find the answer. To answer the problem we need to multiply 2π by 1 or a number that equals to 1 to change 2π into a number that can help us. We need to multiply 2π by 3/3 and it becomes 6π/3 and now we can subtract π/3.

If you are wondering why we multiplied 2π by 1, we did that to turn the number's appearance in a way that would help us without changing the value of the number. 1 is the identity element of multiplication which means that if we multiply any number by 1 or a number that is equivalent to 1, the value or "identity" of the number doesn't change.

The final problem the class worked with is found on the fifth (5th) slide. P(θ) means the point where a ray intersects the circumference of a circle or in this case the unit circle. The unit circle is a circle that has a radius of one on a Cartesian Plain. To find that angle we can use the SOHCAHTOA rule but we need to find the x and y values of the triangle made shown on the slide. The height or opposite side of the triangle is found by looking at the y-coordinate. The x or adjacent length of the triangle can be found by looking at the x-coordinate the given point. Now that we know two sides we can use the Pythagorean Theorem to find the hypotenuse. But wait, if the sides of the triangle that are not the hypotenuse is 8 units long and 6 units long the hypotenuse must be 10 units long because of the Pythagorean triples. Pythagorean triples are side lengths of triangles are whole numbers like 3, 4, 5 and 65, 72, 97. We can now find the unknown angle made by the ray and the x-axis. Now it doesn't matter where we connect the ray to the x-axis creating a right angle the unknown angle is always that angle found. Even if one of the coordinates were to change integers as in 8 into -8. The angle created by the new created ray with the x-axis will always be that angle. The sine, cosine, and tangent of that angle will be the same as the angle's related angle but in some cases the integer might change depending on which quadrant the angle is found.

Related angles are angles that are the same distance in angles from the x-axis. One example would be 30°. Its related angles would be 150°, 210°, 330° because these values are all 30° or 1/12 away from the x-axis π or 2π.

We can find determine the integer of the sine, cosine, and tangent of an angle by the quadrant in which the angle is located. Sine is a measurement of the y-axis which means in the y-value is the 1st and 2nd quadrant it is positive because the y-value is positive and in the 3rd and 4th quadrant the value of sine is negative as with the negative y-value. For cosine it depends on the integer of the x-value. Tangent is sine/cosine which means that if the sine value is positive and the cosine is negative both according to what was said earlier the tangent would be negative because + / - = -. So cast CAST away and remember the sine=y, cosine=x, and tangent=y/x.

That was all we learned today in class. Homework is to watch the Digital Ethics video, read the Digital Ethics post, Exercise #2, and send Mr.K an email to be invited if you haven't yet.

The next scribe will be nelsa!!!

Bye Bye for now and see you in tomorrow's class. Remember "Beans have souls."

Mr.K also told us what new things to look for on the blog in the near future. Such features include a walkie-talkie sort of thing found on the blog which allows people with a microphone on the computer to send voice messages to other people who are online on the blog. Another cool feature is a little number in the corner of the blog which shows the number of people online which helps with using the walkie-talkie feature. The last but not least feature is a chat/shout box which allows us to chat. But with these features also come trust. If we use these features we must follow the Digital Ethics Rules and keep things appropriate. Remember don't post anything that you don't want anybody to know.

Okay onto the math. We did a small review on what we did yesterday. One reminder about units. There are no units for radian answers. All of the units used in finding the radian answer have been reduced to 1 which doesn't change any answer so it is as if they are not there. One more thing, the equation to convert degrees to radians and vice versa is not a formula but a proportion. It shows how degree values are proportional to radian values. So don't memorize that exact equation but remember how degree values are proportional to radian values.

The first question we worked on is on the second slide on the Feb.6 slide post. One solution is showed on the slide but there was one trick that could be used. We know that a full circle is 2π and we can find out that how the hands are found on a clock at 4:00 that it is one-third of the clock or the full circle. So just divide 2π by 3 which gives us our answer with little work.

The second question we worked on is found on the third slide. (Ignore that little message at the bottom at the slide it is irrelevant). That is the correct solution to the problem and I will now explain the process. To answer the problem you must find the area of the circle using the area of a circle equation. Then make a proportional equation like the one on the slide. It shows that 30° out of 360° is equal to the unknown area out of the total area of the circle. Now all that you have to do is solve for the unknown or on the slide, A. Remember to include units where necessary.

That ended our first class for today. Notice that I said first. There was another one.

In our second class we continued with a new problem and the first question required us to know what the word "coterminal" means. Coterminal means end together. So in the question we need to find out what other radian ends where -π/3 (I don't know how to write fractions on the blog) ends or the positive coterminal of -π/3. So we know that a full circle in radians is 2π. we need to subtract π/3 to 2π to find where it ends. If you are wondering what a negative angle is, a negative angle is the same as going backwards from 360° or 2π. One example would be -50°. That is 360°-50°= 310°. Same goes with radians in this problem. We need to subtract π/3 from 2π. By using basic fraction subtracting skills we can find the answer. To answer the problem we need to multiply 2π by 1 or a number that equals to 1 to change 2π into a number that can help us. We need to multiply 2π by 3/3 and it becomes 6π/3 and now we can subtract π/3.

If you are wondering why we multiplied 2π by 1, we did that to turn the number's appearance in a way that would help us without changing the value of the number. 1 is the identity element of multiplication which means that if we multiply any number by 1 or a number that is equivalent to 1, the value or "identity" of the number doesn't change.

The final problem the class worked with is found on the fifth (5th) slide. P(θ) means the point where a ray intersects the circumference of a circle or in this case the unit circle. The unit circle is a circle that has a radius of one on a Cartesian Plain. To find that angle we can use the SOHCAHTOA rule but we need to find the x and y values of the triangle made shown on the slide. The height or opposite side of the triangle is found by looking at the y-coordinate. The x or adjacent length of the triangle can be found by looking at the x-coordinate the given point. Now that we know two sides we can use the Pythagorean Theorem to find the hypotenuse. But wait, if the sides of the triangle that are not the hypotenuse is 8 units long and 6 units long the hypotenuse must be 10 units long because of the Pythagorean triples. Pythagorean triples are side lengths of triangles are whole numbers like 3, 4, 5 and 65, 72, 97. We can now find the unknown angle made by the ray and the x-axis. Now it doesn't matter where we connect the ray to the x-axis creating a right angle the unknown angle is always that angle found. Even if one of the coordinates were to change integers as in 8 into -8. The angle created by the new created ray with the x-axis will always be that angle. The sine, cosine, and tangent of that angle will be the same as the angle's related angle but in some cases the integer might change depending on which quadrant the angle is found.

Related angles are angles that are the same distance in angles from the x-axis. One example would be 30°. Its related angles would be 150°, 210°, 330° because these values are all 30° or 1/12 away from the x-axis π or 2π.

We can find determine the integer of the sine, cosine, and tangent of an angle by the quadrant in which the angle is located. Sine is a measurement of the y-axis which means in the y-value is the 1st and 2nd quadrant it is positive because the y-value is positive and in the 3rd and 4th quadrant the value of sine is negative as with the negative y-value. For cosine it depends on the integer of the x-value. Tangent is sine/cosine which means that if the sine value is positive and the cosine is negative both according to what was said earlier the tangent would be negative because + / - = -. So cast CAST away and remember the sine=y, cosine=x, and tangent=y/x.

That was all we learned today in class. Homework is to watch the Digital Ethics video, read the Digital Ethics post, Exercise #2, and send Mr.K an email to be invited if you haven't yet.

The next scribe will be nelsa!!!

Bye Bye for now and see you in tomorrow's class. Remember "Beans have souls."

Labels:
benofschool,
Circular Functions,
Scribe

Subscribe to:
Posts (Atom)