Tuesday, February 12, 2008


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

1 comment:

Lani Ritter Hall said...

Hi Jamie,

Trigonometric equations, creativity and humor!

What a combination!