Hi everyone, first of all sorry that I can’t be available for a Cover It Live tonight.

Also, I will not be including any cotangent functions on the exam tomorrow – that should be less for you to worry about.

Read the next post as well…

You have determined the period of your sinusoid. Now you need to write an equation. Since the “B” value tells you how many cycles of your transformed function fit into an untransformed version of your graph you can’t just write the period in the equation.
You have to figure out by what factor has the untransformed function been dilated to get the the transformed one.

So, for example, the period of your cosine function is 38 degrees. The untransformed function has a period of 360 degrees. So, by what factor has 360 been dilated to get 38? Essentially 360x = 38. The answer is 38/360 yes? Maybe simplify to 19/180. Now, 19/180 is the
dilation factor but as always with horizontal transformations, what is actually in the equation in the inverse of the transformation. So our final equation is y = a + c[cos 180/19 (theta - d)].

Please spend time connecting what we did in chapter 1 to this current dicsussion – we are doing the exact same thing: you are determining the factor by which the preimage was dilated in order to get the image. Then you are taking the reciprocal of that and putting it into the equation. If the mystery graph was dilated by 2, then you wrote 1/2 in the equation.

What about radians? No difference. Say your sine function has a period of pi/6. By what factor did we dilate 2pi to get pi/6? Essentially 2pi x = (pi/6) and x = (pi/6)/(2pi). This gives us the dilation factor of (1/12) so in the equation we are writing 12.

Work backwards now: if we see 12 in the equation then the dilation factor is (1/12), so (1/12) of 2pi is (pi/6) which is our period.

All of the above holds for tangent and cotangent except you have to remember that untransformed tangent and cotangent have periods of pi or 180 degrees – not 2pi, 360.

Well… Today in class we worked on Problem Set 3-5,  and learned how to compare radians to degrees. To solve most of the problems, you could have used the plate tools we made in class, mainly if you haven’t memorized the radian values. Here’s some examples of the beginning problems to get you started.

60°

To solve this, all you have to know what sixty degrees is equivalent to in radians. If you don’t know the answer off the top of your head, you can refer to your plate, and see that the answer is ?/3.

45°

For this one, you do the same thing you just did for the previous problem, and the answer is ?/4.

We also did problems involving the six trigonometric functions, but in radian mode.

An example of this would be sin2. In Radian mode it would equal .9093.

Another example is tan3. In Radian mode it would equal -.1425.

So that’s all we did in class today.

Hello Honors Pre-Calculus Hour 3! Sorry for the delay in my post, Emily took my scribe because I was absent and I said I would take hers in return…and I subsequently forgot. But better late than never hopefully Thank you for your patience.
At any rate, on Wednesday, December 2nd we worked on the Problem Sets for chapters 3-4 and 3-5.
The next scribe will be…Jake (assuming he hasn’t gone yet and that this counts as my post )

Today we contintinued our paper plate activity from yesterday. We started the day with our tape already folded into 1/3′s and continued to fold our paper until we had 1/12′s. After distinguishing these markings on the tape we then marked them on our paper plates. Our final product of the paper plate is referred to as a unit circle. The image below is similar to the markings we made on the plate.

Source: http://www.mathsisfun.com/geometry/unit-circle.html (This link also has interactive tools that explain relationship between radians and degrees)

This image also shows the calculations of sine and cosine for each radian measure.

After finishing our unit circles we began making the connections with how radians relate to degrees.

60° = ?/3 radians

Since we are still dealing with the same proportions, sin and cos of both 60° and ?/3 radians will be the same.

sin(60°)= 0.866025….

sin(?/3)= 0.866025…..

This also means that the graphs for sin and cos in degrees will appear the same as the graphs for sin and cos in radians. The only difference will be the horizontal scale.

Homework for tonight: Read and fill out yellow sheets for 3-4 and 3-5

*Exam Next Tuesday

The next scribe will be Emily