As I was on the computer one night I was looking for something that would help me complete the table where we were given a measurement in degrees or radians and we had to solve for sinx, cosx, and cotx.  I found this cool website that has all of the 6 trig functions and lets you try and figure out the conversions of degrees to radians and what they equal when they are put into a trig function.  So for any of you that struggled with this like I did, here is a website that is a great study tool for the final exam.

http://www.studystack.com/studytable-3602

After another long week, the class was anxious for the weekend. But it wasnt quite here yet. As we entered the classroom, Mr Bienek had a problem on the smart board. We had alot of material to cover and Mr. B encouraged us to get started.

The problem stated 1.) Find the measure of the angle a and 2.) what other angles share a vertical component of .7

After discussing the problem we were asked many questions regarding the vertical coordinates and the many different angles in the circle. We came to the conclusions that…

~How do we get the vertical coordinate?
sin (a) = .7
sin^-1 (.7) = a
a = .78 . . .

Big Angle= 180 – a = pi – a
and sin^-1(.7) = a = pi -sin^-1(.7)

Next we solved for this equation regarding the circle:

Solve: 10sin ( x-.2) = -3 domain: [0,4pi]

1.) sin (x-.2) = -3/10
2.) sin (x-.2) =sin^(-3/10) or pi – sin^-1(-3/10)
3.) x = sin^-1 (-3/100 = .2 or ( pi – sin^-1 (-3/10))= .2
4.) Solve on Calculator
x = -.1046 . . . or x 3.6462 . . .

(~ Add one full revolution to find solutions.)
No stretch so period = 2pi = 1 full revolution

x = 6.1784 . . . x = 3.6462 . . .
x = 12.4616 . . . x = 9.9294 . . .

Next we had a short discussion on the the angles that would be fair game for the test. They should be memorized by now…but just in case here they are.
sin^-1 (( square root 3) /2) = pi/3 = 60 degrees
sin^-1 (1/2) = pi/3 = 30 degrees
cos^-1 (1/2) = pi/3 = 60 degrees
cos^-1 ((square root 3) / 2 ) = pi/6 = 30 degrees
~ We need to be familiar with these ratios !

Next we discussed the cosine of a circle…

and then moved on to Tangent…

The tangent problem was to 1.) Use tangent to find the measure of angle b. 2.) what other angles are possible solutions?

1.) tan b = sin/cos = .43/.90 = .48
tan^-1 ( .48) = .4457 . . .
pi + tan^-1 ( .48) = tan ^-1 9 .48) = .4457 . . .

~ two possible places on the unit circle.

Lastly we had to solve in Degrees this equation:

4 tan ( 2x ) = -5 Domain (-45, 180 ]

tan 2x = -5/4
2x= tan^-1 (-5/4) or 180 + tan^-1 (-5/4)
x = ( tan^-1 ( -5/4))/2 or 180+ (tan^-1 ( -5/4))/2
x = -25.670 . . . or x = 64.3299 . . .
~ To find all possible solutions add 90. ( 90= period)

Well that was our class on friday complete with the smart board lesson. Make sure to study how to deal with tangent cosine and sine .

Our homework was to do problems in 4.4 #15-20.

The next scribe will be…Kellie.

Mr. B wants me to experiment with some HTML code today, so enjoy the text effects!!

Today in class, we started off with a nice relaxing pause (with a shocking lack of music) while Mr B. figured out who would do the presentations for 1-3. I was personally preparing the problems I submitted to present, as I saw many others doing. We were then reminded that we should be keeping track of how many we’ve claimed. Emily led off our claims with problem #5,followed by Amy with problem #12, Grant with problem #14, and Jessica with problem #19.

~~Emily (#5)

Emily demonstrated that the way to put the equation we were given in terms of x is g(x)=1+sq(9-(x/2)². She then explained that the g(x)graph is an x dilation of 2 and a y-translation of 1 compared to the preimage and finished up with showing us her graph.

~~Amy (#12)

Amy spent a bit of time setting her problem up, drawing the graph to make her explanation easier. She explained that the image of her graph was translated down by 4 and dilated on the x-axis by 3 from the preimage. She then correctly gave us her equation of the image, g(x)=f(1/3x)-4, explaining that 1/3 resulted in a dilation of 3 because of the “backwards world” properties of parentheses. The class was confused as to what in the world she was talking about, but applauded regardless.

~~Grant (#14)

Having an amazing worldly presentation to follow, grant made sure that everything was as aesthetically pleasing as possible, down to the orange color of his cyber calculator, and the dottedness of his lines. Grant seemed a bit unsure of his problem, but proceeded this mimentary slip with a very nice presentation. His first equation on his graphing calculator was Y₁=4sq(1-x/3)+2.25(x-1)-4, explaining about the fabled opposite world we’ve all grown to know and love. His graphs had slight problems getting just the way he wanted them, but with the help of the class he got through. With the addition of the boolean variables his graphs were complete, and they correctly resembled
those in problem #12. His presentation was thusly concluded.

~~Jessica (#19)

With the last presentation to go, and many tough acts to follow, Jessica set up. Her graphs were constructed, both image and preimage. Her equation for the image was g(x)=5f(1/9x), because the y-dilation is 5 and the x-dilation is 9 (backwards world… Surprise, right?). Her presentation was short and to the point, but accurate nontheless.

And so the presentations concluded. Mr. B passed out a sheet on lesson 1-4, and challenged us to think of an equation that relates median age to the percent of that group that has a fear of flying. Kari had the idea of forming a “stat plot” on her calculator. We proceeded to do so on all of our calculators, and after a minute or two of plastic clicking noises, we were done. We concluded that there is clearly a relationship, but not a linear one. Someone suggested that we make it linear by squaring the independent variables, citing a physics lesson for the idea, and lo and behold, it worked!!! To end our lesson, Mr. B challenged us to come up with a way besides the one we used in class to make the graph linear. As the bell rang, we all trooped out, amazed that we actually used something from one class in another. Who’d have known this stuff was useful??

I hope you liked my post today, and the changes that HTML code can bring. With Mr. B’s permission, tomorrow I’ll post a short tutorial that outlines the main ways to change and edit text using HTML. It’s not hard. And sorry if some of the stuff on here either doesn’t work or looks weird, I was experimenting and I’m not sure how much code this site will accept, and I copy-pasted this from a notepad document, not using any of the shortcuts on the blog site. If anyone would like to see what my source code looks like, just ask me in class. And tell me what you liked and didn’t like with comments!

Tomorrow’s scribe will be Shelby. Have a good day, and happy commenting!!!

~~~~~~~~ANDY~~~~~~~~

Today we started class with presentations Phil showed us how to do number 5, ALyssa did 12, I did 14 and Matt did number 19. Everyone did a wonderful job with explaining their problems.

After that we started a problem that used facts about the percentage of people who airplane fears at a median age. First, we plotted the point on our calculater. To do that we used stat plots. (Stat-Edit) Then we entered the ages in for L1 and percentages in for L2. After all your points are entered you can graph it. To do that you have to go to 2nd then Stat Plot (y=) Then you go into plot 1 and turn it on and make sure your xlist is L1 and ylist is L2. The window for this graph was about xmin20-xmax75 and ymin0-ymax20. The graph of these points looked like an exponetial graph.

So then in order to find an equation for the points we made the graph linear. To do that we changed the x values to x^2. Back in the Stat Plot list you use your L3. And put in the equation “L1″^2. and it will square all teh x values. Next you go back to the Stat Plot (2nd- y=) then in the first instead of your xlist being L1 you make it L3. And they ylist is still L2. The you graph it again. To get a window that wooks good you can use zoom and then 9, which will use the stats to fit the window. This graph is linear!!

Homework for the night was to try and find another way to make the orginal curved line linear in a different way.

The next scribe will be……Rose =)

Today was a fairly good Monday Precalc class. We reviewed the old work and learned some new things.

First we started class with Mr. B explaining more about inverse variation and direct variation functions from the Restricted Domains and Boolean Variables worksheet. We learned how to restrict the domain of a function on the calculator. It is surprisingly easy to do. After you have finished typing in the equation you will type a division sign and and in parenthesis type this domain using symbols in the test category of your calulator. If there are two parts you will type both in the parenthesis with using and to split them up. And can be found in the logic category, but and is only true when both restrictions are true. It is false when one is not part of the domain is not true. When the domain is true, the entire equation is divided by 1, not changing it, but when the domain is false, the equation is divided by 0 giving you an irrational answer according to Boolean variables.

Once you have typed the restricting domains in, you can graph the equations. This is very helpful when you only need to see one part of the entire function. Not only can this be done on your calculator, but it also works with winplot. You are able to change the low and high of the domian.

The second part of class was spent looking at functions and finding ways of moving, stretching, and compressing them. Chapter 1:3, our homework, was all about these changes.

Graphs can be moved around on a grid by adding or subtracting from the original function in a new one. Say you want a graph to move up 3 units on the graph. To do this you would at 3 to the function. It might look like this as seen on the worksheet:

g(x) = f(x) + 3

As you can see, the number 3 is on the outside of the f(x). This means whatever the y value for the f(x) is added to 3 to produce the g(x).

The opposite is done if you want to move the graph right or left. To do this you need to add or subtract a value on the inside of the f(x). Doing this directly changes the x value instead of the y value when moving up or down. If you wanted the graph to move 3 units to the right this is what the function would look like:

g(x) = f(x-3)

If this is confusing a good way to think of this is by pretending to move the origin. The following two types of changes are translations. Dialations can also be made that stretch or compress the graph. These are made by multiplication inside and outside the parenthesis of a function.

To stretch, or dialate, the graph vertically you would multiply the original function by the number you wanted to dialate it by. To continue with examples on the worksheet I will use the number 2.

g(x) = 2f(x)

Whatever is the value of y is multiplied by 2. On the contrary, dialating horizontally is again done inside the parenthesis. Instead of multiplying by a whole number you will multiply by the reciprocal of the number you want to dialate by.

g(x) = f(1/2 x)

Here is a recap of today’s lesson:

1. Restricting domains can be used on a graphing calculator and winplot.

2. Domain is true = divide equation by 1, Domain is false = divide equation by 0

3. Translate graph by adding and subtracting inside and outside parenthesis of original function.

4. Dialate graph by multiplying by whole numbers and fractions inside and outside parenthesis of original function.

Hope this helps clear up what we learned today. The next scribe will be Allison.

-Kendall