Precal Blog

Today in class, we began by going over the claims for sections 2-2 and 2-3.

Ali presented the first problem and it asked for the refence angle if theta=-2746 degrees.

What he did was divide -2746 by 360, and the answer ends up being -7.62777778.  THis number means that theta makes 7 full rotations in the negative direction and a partial rotation of .62777778.  Net he multiplied -.62777778 by 360, and that answer would give him the measure of theta so he could plot it on a uv axis without spiraling seven times around the origin.  The value for theta is -226 degrees, making it in the 3rd quardrant in the negative direction.  Then to find the reference of theta,  Ali completed the rotation to the nearest horizontal axis, keeping in mind that it must always go in the positive direction.  It would be drawn to 180 degrees, and 226-180= 46 degrees.

Jared presented the next problem, and it asked to find the reference angle for an angle measuring 352 degrees 16′ 44″.

 The first step was to draw the angle in the fourth quadrant, because it was between 270 and 360.  Then the reference angle can be drawn to the nearest horizontal axis.  But since the initial angle is in degrees minutes and seconds, you need to subtract it from 360 degrees 0′ 0″ properly.  The way to do it is to just simply borrow from the previous number.  The answer comes out to be 7 degrees 43′ 16″.

For this problem, Pat was asked to write g(x) when the pre-image is f(x).

 Visually you can see that it is dilated in both the x and y direction.  To find out the value for the vertical dilation, you take the y point of one of the humps of the pre-image and compare it to the y coordinate of one of the humps of the image.  In this case, the peak oif the pre-image hump is 3, and the peak of the image hump is 9, making it a dilation by 3.  g(x)=3f(x).  For the horizontal dilation, count the horizontal distance between the humps of the pre-image and and compare.  The pre-image is twice as small as the image, meaning that the image got two times as big.  g(x)=3f(1/2x).  It is 1/2x inside the parenthesis because it is always inverse in the x direction.

In this problem presented by Streim, he was asked to find the reference angle for 300 degrees and calculate the sine and cosine of the initial angle and reference angle.  It is the same as the first two problems that were presented in that you need to find the reference angle.  The reference angle ends up being 60 degrees.  We have the sin and cosine value memorized for 60 (sq. root(3)/2=sin60 and 1/2=cos60) so we know that since 60 is 300’s reference angle, their sines are opposite and their cosines are equal.

Justin’s problem gave him a point (-7,-24) and it told him to find the sine and cosine of its angle.  First he needs to draw the triangle, knowing the veritcal side is -7 and the horizontal side is -24.  Then use the Pythagorean theorem to find the hypotenuse, and use sine and cosine functions to solve for it.  Answer: sin=-7/25, cos=-24/25

For this one, Jocelyn needed to graph the given equation and tell its transformations.  4cos tells you it is dilated vertically by 4 and (theta+60) means it is translated 60 degrees to the left.  By the way the drawing was just superb for this one hahaha.

After claims, Mr. Bieniek talked abou the difference between inverse and reciprocal.  The notation is very slightly different but is critical.  For inverse, the notation is cos^-1(theta).  For reciprocal the notation is cos(theta)^-1.  He also talked about how theta and the inverse function (sine, cosine, tangent) of theta are the same. 

The next scribe is Stinz Leerix, who is droppin’ a new mixtape ‘Tha Letdown”.  Coming soon to stores near you

Today in class we learned about Reciprocal Trig. Functions and how they are related to their Reciprocals. To start out, our original functions are Sin of Theta = V/R, Cos of Theta = U/R, and Tan of theta = V/U. Sin’s reciprocal is CSC theta = 1/sin of Theta = R/V. Cos’s reciprocal is SEC theta = 1/Cos theta = R/U. Last, Tan’s reciprocal is COT theta = 1/Tan theta = U/R. It is very important that everyone is clear that the Reciprocals and Inverse are two DIFFERENT things. Also, remember that 1/csc = Sin theta, 1/sec = Sin theta, and 1/cot = Tan theta.

Next we moved onto an angle that contains the point (-4,-6). Sin theta = -6/sq root of 52 and csc = sq. root of 52/-6. Cos theta = -4/sq. root of 52 and sec =  sq. root of 52/-4. Tan theta = -6/-4 =3/2 and cot = 2/3 (tan is positive in the 3rd quadrant).

Lastly we found the value for all 6 trig. functions for theta = 330 degrees. Sin(30) = -1/2 and csc = 2/-1. Cos(30) = sq. root of 3/ 2 and sec = 2/sq. root of 3 (cos is positive in the 4th quadrant). Tan(30) = -1/sq. root of 3 and cot = sq. root of 3/-1.

As a side note know that, in the first quadrant all trig functions are positive, the second quadrant has only sin positive, the third quadrant has only tan positive, and in the forth quadrant only cos is positive.

PS, the next scibe is Dushan

We began class today on the packet with the final example question:

The graph of points from (U,V). U= -8 and V= 5, comprise in the formation of a 90 degree angle. From this we can use Pythagorean theorem a^2+b^2=c^2. (the reference angle is counterclockwise and is positive but is not needed for this). This answer we found  from the terminal position’s length of square root of 8.  Looking back and remembering that sine= opposite / hypotenuse and cosine = adjacent/ hypotenuse and later finding the sine and cosine to the terminal position. Sine theta = 5/square root:89 and cosine theta = -8/square root:89. Now since the terminal angle ends in the second quadrant coordinate plane, cosine is going to be (-) and sine is going to be positive (+).Towards the end of class we looked at graphs of the sine and cosine functions (graphs A and B). Graph A is y= cos(X) and Graph B is y= sin(X). In comparing the sinusoids one can see the periodic function as cos(X) repeats itself through a cycle that sin(X) is inversly related to.

The next scribe is Jack K.

Today, the class started out with a packet. The packet had tables and picture in it, and we took notes on it. If you were sick, it would be hard to follow this since you don’t have the packet. So, hopefully the pictures will help you out. The post is image loaded, which translates into fun!

What we learned: right triangles, sine and cosine, relationships between sine and cosine, and more! All of this on the (u,v) plane, which is exactly like the (x,y) axes except the (u,v) is used for angles, for some reason. 

So, from the image we learn what the sine and cosine of “theta” is, in a right triangle. No need to delve into more details because this should be a review of what we learned last year. If it’ll help you out, remember SOH CAH TOA. 

In this image, we review the special properties of a 30-60-90 triangle. The leg opposite of  the 30 degree angle is always half of the hypotenuse because if you think of the 30-60-90 triangle as half of an equilateral triangle, that leg would be half the side length for the equilateral triangle. Then use the Pythagorean Theorem to solve for the remaining side, which is the square  root of 3/4.   

If you notice, the sine of 30 and the cosine of 60 are equal. Also the sine of 60 and the cosine of 30 are equal. This is because the “co” from cosine is the same as the “co” from complementary. The sine of an angle and the cosine of its complementary angle will always be the same. (Complementary = two angles that add up to 90) Mathematicians discovered this relationship and named the “co-of-sine” as cosine because of the relationship, if you get my drift.

*Note: You don’t need to rationalize fractions. It is unnecessary except on college entrance exams. Only reason why people did it back in the old days was because they didn’t have a handy dandy calculator, and dividing by an irrational number they didn’t know the exact value of was…. icky. -according to reversearp

As you can see, a 45-90-45 triangle has a hypotenuse of the square root of 2 times the length of the other two sides. And the cosine and sine of 45 are equal to each other because they are both complementary to each other.

Okay, so in this image comes the most important and loaded statement of the class period. In this picture, “theta” is put in standard position which I believe is just the first quadrant. Now, here comes the big statement. The value of “u” for  the point “r” is just the length of the adjacent leg to “theta.” And you associate it with the cosine because the cosine of “theta” is adjacent/hypotenuse. And the other part of the statement is that the value of “v” is the length of the opposite side. Also sine is associated with “v.” When it all comes down to it, the sine of “theta” is v/r and the cosine of “theta” is u/r.

*reversearp challenge- try to do something mathematically in order to get the tangent of “theta” = v/u. Participation not required.

Now you’re asking how do we solve for the sine and cosine of angles over 90 degrees? Well, to do that we need the reference angle! The reference angle of 120 happens to be 60 degrees. (Remember: reference angles are made to the closest horizontal axis) And it’s located in the second quadrant, which is important. If you remember, the reference angle and the original angle have the same magnitude (value). So, the magnitude of the cosine and sine of 120 would be the same as the cosine and sine of 60 degrees, respectively.

But what about the direction (which quadrant it’s in and whether the values are positive or negative)? Well, since the reference angle terminates in the second quadrant, the cosine is negative and the sine is positive. Why? Because “u” values are associated with cosine and “v” values associate with sine. And the quadrant in which the reference point terminates in determines the “u” and “v” value’s negativeness or positiveness. If you still don’t understand, look at the image above for a clearer picture; it also shows the value of the quadrant. First one is (+,+), second: (-,+), third: (-,-), and fourth: (+,-).

Example time

Reference angle of 225 is 45 degrees. Take sine and cosine values of 45 and set them as the sine and cosine values for 225. Determine which values are negative or positive. By looking at the image, reference point terminates in third quadrant, in which both “u” and “v” values are negative, so both sine and cosine values are negative.

Reference angle of 330 is 30 degrees. Sine and cosine values of 30 degrees same as 330 degrees. Original angle terminates in fourth quadrant, so “u” values are positive and “v” values are negative, which in turn makes the cosine value positive and sine value negative.

Overview: we learned a lot of stuff. As long as you remember the rules for the 30-60-90 triangle and 45-90-45 triangle, the remaining “stuff” shall come easy.

So today in class we stoped learning about funtions and reflections and what not and learned about angle measure rotations.  We first started out by being informed of the different parts of the graphs.  The graphs used to do these use a UV coordinate plane instead of a XY.  In this diagram the initial side (the side your starting from) is green and the terminal side (the ending line) is red.  The degree measurement is called Theta (“the ta” aka ?).

When making these graphs you must remember that

• the vertex is the origin
• the initial side line lies on the posi side of the U axis
• Posi angles rotate counter clockwise

more examples of graphs are

When two angles terminate in the same place, like below, they are conterminal

Next we learned about reference angles.  These make it easier to calculate angles because you only have to deal with quadrant one. The reference angle is represented as ?ref. When angles are greater then 90 degrees, they can be represented by a reference angles.  Reference angles and the original angle both have the same magnitude aka the sin(133)= sin(47).

When making these remember

• They are drawn between teh terminal side and the U axis
• They are always posi

Thats it. Thanks team(:

Ok, so today we had on claims from 1-6. The problems available for claims were 1, 8, 9, and 11. Thank you Sara and Scott for presenting today. Im sorry for my lack of pictures.
  Sara presented problem 1 for us todayFor problem 1a., -f(x), it is a reflection of the pre-image across the x-axis making all negative y values positive and vice versa. For 1b., we were asked to find f(-x). This reflects the  pre-image across the y-axis making all negative x values positive and vice versa. Problem 1c. asks you to find |f(x)|. This makes it so that all the negative y-values from the pre-image become positive but all positive y-values stay the same on the graph. Problem 1d. asks you to find f(|x|). In doing this, all the points of the pre-image in the I and IV Quadrant stay the same, because all of their x-values are positive. In quadrants II and III, you find the absolute value of all the x-values of the pre-image. Then you take the |x| values and plug them into the f(x) equation. After doing this, the image should look somewhat like a parabola. The points in the II and  III quadrant will be a reflection across the y-axis of the points of the pre-image.
Sadly noone claimed problem number 8, so our class moved onto problem 9 which was well explained by Scott, but caused some confusion for some people including myself, so I do not want to try and explain this problem because I am not sure if how I am explaining it is correct.
  Because of lack of time, at the end of class we were told that we would work on the claim for problem number 11 tomorrow.
The next scribe will be………………………Jocelyn.

Alright, so today we started a new section, 1-6. First off we reviewed what happened to a graph for each of the following functions:
f(x)=f(-x): the graph flips over the y-axis
f(x)=-f(x): the graph flips over the x-axis
f(x)=|f(x)|: all negative y values become positive
f(x)=f(|x|): the positive values of x stay the same while all negative values for x are eliminated. Then the positive values for x are reflected over the y-axis
*the exploration 1-6a worksheet gives examples of all of these

After that we started talking about symmetry in a graph.We learned that there are two types of symmetry- odd and even- and then we looked at six graphs to determine if they were symmetrical.

Two of the graphs had even symmetry. So f(-x)=f(x). This means that the graph flips over the y-axis and that there will always be two x values that have the same y value.

Two more graphs demonstrated odd symmetry where f(-x)=-f(x) and the graph was reflected over both the x and y axes.

After we identified the graphs that were symmetrical and the types of symmetry used in the ones that were we learned how to algebraically determine if a graph was symmetrical and if it was, if it had even or odd symmetry.

We can figure out if the graph has even or odd symmetry by applying f(-x) and –f(x) to the f(x) equation and seeing which two are equal. If f(-x) and f(x) are equal then the graph has even symmetry, if f(-x) and –f(x) are equal then the graph has odd symmetry. If none of these functions are equal then the graph has no symmetry.

The first problem we did was this one:

We know that this graph has odd symmetry because f(-x)=-f(x).

The second promblem was

Again, we know that this graph is odd using the same rules from above.

The third example problem we had didn’t have symmetry at all:

When you look at each of the functions you’ll see that none of them are equal to each other meaning that it is neither even or odd symmetry.

The final example problem:

Quite a few people got confused with this one because at first glance none of the function were equal to each other. In order for us to see that there was really even symmetry we had to multiply our simplified version of f(-x) to see that it was the same as the equation for f(x).

At the end of class we were assigned problems and told which ones were available for claims.

The next scribe will be Shannon

Alrighty so we started off class by talking about new types of transformations. We focused on what happens when f(x) = f(-x), f(x) = -f(x), f(x)= |f(x)| and when f(x) = f(|x|). When f(x) = f(-x), the graph flips over the x-axis. When f(x) = -f(x) the graph flips over the y-axis. When f(x)= |f(x)| all of the y values are made positive. And finally when f(x) = f(|x|) the graph where x is positive is flipped so that when x is negative it equals the same. After that we began discussing symmetry in a graph. We then looked at 6 graphs and determined if they were symmetrical. We also learned that there are two types of symmetry, odd and even. Even symmetry is when (f-x) =f(x) and it is a flip across the y axis and x equals the same. Here are examples that we looked at

As you can see with both of these the y values are the same if x is positive or negative. That is because f(x) = f (-x). So if x is positive or negative y always equals the same. These graphs demonstrate even symmetry.

The graphs that showed odd were

In these f (-x) = f(x) so as you can see the y values or OPPOSITES. These are reflections across the y AND x axes.

It is important to remember that there are plenty of graphs that are not odd or even and that reflections across just the x axis aren’t functions because then you would have multiple y values for a single x value.

Next we looked at how to algebraically determine if graphs are odd or even.

Here’s a picture of the 1st problem we did

We know the graph is odd if f (-x) is -f(x). To do this you simply plug into the equation and simplify and see if you get the same results. In this case you did.

Here’s another example of odd which tripped quite a few people up

The important thing to remember in this is with fractions you are only multiplying into the TOP part (since in essences you are multiplying by ).

Here.s an example when the graph is neither odd nor even

And now for the final problem…

This one is even and caused some problems for people. It is important to remember that you can factor out of your final result. By doing this you actually see that the two equations are equal.

At the end of class we were assigned 1-6 and told the claims would be problems 1, 8,9,11 but that we should NOT just do the claims.

The next scribe will be Adam K

Hi all – sorry I didn’t post this sooner but your precal book is available on-line here: http://www.keymath.com/PreCalc.

You will need a “ClassPass” in order to access the entire contents of the book so you’ll have to send me an email to get it. I carry my ipod touch with me most of the time now so you should get a pretty prompt reply.