## Tuesday, October 9, 2012

### "Memorizing" the Unit Circle

I, like many others I'm sure, was told that I needed to memorize the unit circle while studying Pre-calculus in High School. This looks like a very daunting task at first glance, and it caused me a lot of frustration the first time I learned it.

I was taught to memorize them straight. I don't know why, because there's a much, much, better way to go about it. I'm going to start with the full unit circle and widdle down "memorization" to the bare minimum. Once you can start to see the reality of what the Unit Circle is, it becomes way easier to interpret. The Unit Circle
First off, the unit circle is just a circle of radius 1. Knowing this, we can determine that its circumference is $c = 2\pi r = 2\pi 1 = 2\pi$ radians. Points on the unit circle are easy to find knowing that the circumference is just $2\pi$. We just have to figure out what fraction of $2\pi$ our target angle is, and locate that section on the circle.

So what's our goal? We need to be able to extract the trig functions from the unit circle:
$$sin\theta, cos\theta, tan\theta, csc\theta, sec\theta, tan\theta$$
at each interval of $\frac{\pi}{6}$, or $30˚$, and each interval of $\frac{\pi}{4}$, or $45˚$.

To do this, we need to know a little bit about the way that the coordinate system on the unit circle works. At any given point on the unit circle, we can make a triangle starting from the center of the circle to said point that rests on the x axis.  The hypotenuse of this triangle will always be $1$, simply by the definition of the unit circle. We know that $sin = \frac{opposite}{hypotenuse}$ and $cos = \frac{adjacent}{hypotenuse}$ for any angle on a triangle. We have noted that our $hypotenuse$ will always be $1$, so the $sin$ of any angle on the unit circle will simply be the $\frac{opposite}{1}$, or the length of the opposite side, otherwise known as our $y$ coordinate. Similarly, $cos = \frac{adjacent}{1}$, or the length of the adjacent side: our $x$ coordinate.

Knowing this, the cases of $0\pi, \frac{\pi}{2}, \pi,$ and $\frac{3\pi}{2}$ become trivial -- All we need to do is point out which coordinate is $0$, and which one is $1$, and we have both $cos$ and $sin$.

So, what about the rest of them? These are the only three fractions that I like to keep in mind when dealing with the unit circle angles. Commit these to memory:
$$\frac{1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}$$

These (and their negations) will be our $cos$ and $sin$ values for any "special" point on the unit circle, and as such, they are all we need to remember (the rest of the trig functions can be derived from these values).

Let's first just deal with the first quadrant angles, $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$. The first thing that I want to do is order the fractions above from smallest to largest. I've actually written them in the correct order; so I won't bother writing them out again. Now, let's think about this for a second. In the first quadrant, out of our three points, where is $sin$ (our $y$ coordinate) the smallest? We start at $0$, and our $y$ value gets closer and closer to $1$ as the angle increases. This means that the smallest angle of our three ($\frac{\pi}{6}$) gets the smallest $sin$ value. Following this convention, we can say that our middle angle ($\frac{\pi}{4}$) gets the middle $sin$ value ($\frac{\sqrt2}{2}$), and our largest angle ($\frac{\pi}{3}$) gets the largest $sin$ value ($\frac{\sqrt3}{2}$).

But wait! Didn't we say that our $cos$ value is simply the $y$ coordinate? Yes, and as such, this idea can be applied to the $cos$ values as well. $cos$ ($x$) is decreasing as we move from $0$ to $2\pi$ on our unit circle, though, so as angles increase, $cos$ decreases. Apply the logic from before, and we have:
$$cos \frac{\pi}{6} = \frac{\sqrt3}{2}, cos \frac{\pi}{4} = \frac{\sqrt2}{2}, cos \frac{\pi}{3} = \frac{1}{2}$$

Knowing this, you know the rest of the values on the unit circle. Why? Because we can see where the $x$ or $y$ coordinates are negative, and it is obvious which one is larger than the other at any point. For example, take $\frac{5\pi}{6}$. Here, we can see that the $x$ coordinate is negative, but the $y$ coordinate is positive. So, we have a positive $sin$ and a negative $cos$. The length of the $y$ side is clearly shorter than the length of the $x$ side, so we can see that $sin \frac{5\pi}{6} = \frac{1}{2}$ and $cos \frac{5\pi}{6} = -\frac{\sqrt3}{2}$. Note: At intervals of $\frac{\pi}{4}$, we will always have the same ratio for both $sin$ and $cos$: $\frac{sqrt2}{2}$. It's just a matter of getting the signs correct!

So, what about the rest of the trig functions? Well, let's start with $tan$. We know that $tan\theta = \frac{opposite}{adjacent}$, right? Well, our $adjacent$ side on the unit circle is just $cos$. Our $opposite$ side is just $sin$. Therefore, on the unit circle, $tan\theta = \frac{sin\theta}{cos\theta}$. Since we can get $sin$ and $cos$, we can just divide them out and get our $tan$ value. Simple.

$sec\theta$,   $csc\theta$,  and $cot\theta$ then become trivial. All we need to do is take the reciprocals of their counterparts ($cos\theta$, $sec\theta$, and $tan\theta$ respectively). $\frac{1}{2}$ becomes $2$, $\frac{\sqrt2}{2}$ becomes $\sqrt2$, and so on.

And that's all of the trig functions for the "essential points" on the unit circle. All that is truly necessary is the recalling of three simple fractions and a few key concepts, and the whole thing becomes easy.

Hope this helps!

-Ben