Precal Blog

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.

Interesting problem today, converting a vertically stretched base-10 log into a base-n log with no vertical stretch. Does it work in general? What does it look like if you try to convert a dilated base-3 log to a non-dilated base-n log?

We also talked about the two forms of the logistic function and showed them equivalent although a few people thought it was magic.

I decided on the Cramer’s rule homework because I think it is interesting (the rule I mean) and something you will definitely see again in your college studies. It also extends the matrix concepts we talked about. I’d like that by next Friday.

What do you do if you want to model exponential growth but there is a constraint as to how much growth there can be (say in a population that will eventually exhaust its habitat)? You use a logistic function…but more on that tomorrow.

Need an additional take on AP History or Physics? The internet archive has archived complete AP History and Physics courses. They are all multi-media and very well done. It is surprising that more people don’t know about them and utilize them.