A unit circle is just a damn circle with a radius of 1. That’s it, baby! It’s a circle with a radius of 1 that let’s us calculate trigonometric functions like nothing. Y’all remember what trig functions are all about. They’re functions that tell us a whole hell of a lot about triangles. The trick, though, is to be able to know what the ratios for any given angle are. If you know that, then you’ll be able to use them to do all kinds of shit. The unit circle makes it easy to visualize how the ratios behave. Let’s get into it.

## Draw the thing

All you need to do to sketch out a unit circle is draw a circle that has a radius of 1, and center that bitch at (0,0).

Once you have the circle, you can start drawing out triangles by starting at (0,0) and drawing a line out to the edge of the circle.

We know that this line has a length of 1, because that’s how we drew the fucking circle. So then the hypotenuse of the triangle you drew is 1, and let’s say for now the sides are $x$ and $y$. The last thing we need to worry about is the angle between the $x$ axis and the line you drew. Call that bitch $\theta$. So now we have a right triangle with an angle, an adjacent side, an opposite side, and a hypotenuse.

So we know that

$$\sin(\theta)=\dfrac{x}{1}=x$$

$$\cos(\theta)=\dfrac{y}{1}=y$$

$$\tan(\theta)=\dfrac{y}{x}$$

So all we’re really looking for are the coordinates of that point we drew on the circle $(x,y)$. With $x$ and $y$ we can calculate all the trig ratios for $\theta.$

## Add values & shit

Now, it just so happens that we know what the measurements of $x$ and $y$ are for certain angle measurements. Isn’t that just fucking convenient? For 30° $x$ is $\dfrac{\sqrt{3}}{2}$ and $y$ is $\dfrac{1}{2}$.

Meaning that $\sin(30^\circ)=\dfrac{1}{2}$ and $\cos(30^\circ)=\dfrac{\sqrt{3}}{2}$

For 60° $x$ is $\dfrac{1}{2}$ and $y$ is $\dfrac{\sqrt{3}}{2}$.

Meaning that $\sin(60^\circ)=\dfrac{\sqrt{3}}{2}$ and $\cos(60^\circ)=\dfrac{1}{2}$

For 45° $x$ is $\dfrac{\sqrt{2}}{2}$ and y is $\dfrac{\sqrt{2}}{2}$.

Meaning that $\sin(45^\circ)=\dfrac{\sqrt{2}}{2}$ and $\cos(45^\circ)=\dfrac{\sqrt{2}}{2}$

Angles like 0° and 90° are on the axes so they are just combinations of 1s and 0s.

So $\sin(0^\circ)=0$ and $\cos(0^\circ)=1$, $\sin(90^\circ)=1$ and $\cos(90^\circ)=0$, and so on and so forth.

These are obviously only values in the top right corner aka the 1st quadrant. This works for angles between 0 and 90 degrees, but what about angles higher than 90 degrees? Don’t trip fam… it’s just a mirror image of the same shit. It’s the same triangles just copy pasted into the different quadrants. Por ejemplo, 135° is just 90° plus 45°, so you can just use the same 45° triangle values transposed over to quadrant 2. Giving us $\cos(135^\circ)=-\dfrac{\sqrt{2}}{2}$ and $\sin(135^\circ)=\dfrac{\sqrt{2}}{2}$.

You can pull the same shit for an angle like 210°… just 180° plus 30°… $\cos(210^\circ)=-\dfrac{\sqrt{3}}{2}$ and $\sin(210^\circ)=-\dfrac{1}{2}$.

So you really know the trig ratios for any angle that is 30, 45, or 60 degrees away from any of the axes. Just make sure to get your positive/negatives right depending on what quadrant you’re in.

Sidenote: I’m not really calculating tangent ratios because it’s just the ratio of sine to cosine. For example:

$$\tan(30^\circ)=\dfrac{\sin(30^\circ)}{\cos(30^\circ)}$$

$$\tan(30^\circ)=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$$

$$\tan(30^\circ)=\dfrac{1}{\sqrt{3}}$$

$$\tan(30^\circ)=\dfrac{\sqrt{3}}{3}$$

Simple enough.

## Use it

### Example 1

Let’s say you have some right triangle, and you know one of angles is 30 degrees and the side adjacent to that angle is 10 inches. Yes I’m using Imperial AKA Freedom units. Come at me. Let’s call the opposite side $a$ and the hypotenuse $b$.

Motherfucker you know everything about that triangle. Just replace $\cos(30^\circ)$ with $\dfrac{\sqrt{3}}{2}$ and you’re good to go.

$$\cos(30^\circ)=\dfrac{10\,\text{in}}{b}$$

$$\dfrac{\sqrt{3}}{2}=\dfrac{10\,\text{in}}{b}$$

$$\dfrac{b\sqrt{3}}{2}=10\,\text{in}$$

$$b\sqrt{3}=20 \, \text{in}$$

$$b=\dfrac{20\,\text{in}}{\sqrt{3}}$$

$$b=\dfrac{20\sqrt{3}\,\text{in}}{3}$$

At this point you could use either $\sin(30^\circ)$ or $\cos(60^\circ)$ to figure out a. They’re identical so it really don’t matter. Alternatively, we could use tangent since that uses the 2 sides instead of the hypotenuse ($\frac{opposite}{adjacent}$), and we know one of the sides. Let’s do that. We already found the value of $\tan(30^\circ)$ so it should be simple.

$$\tan(30^\circ)=\dfrac{a}{10\,\text{in}}$$

$$\dfrac{\sqrt{3}}{3} = \dfrac{a}{10\,\text{in}}$$

$$\dfrac{10\sqrt{3}\,\text{in}}{3}=a$$

I’m not going to calculate that monster because I’m above that there I said it. Anyway, we figured out all the sides of the triangle. They’re ugly as sin but we did it. Look at you doing math!

### Example 2

Or say you have a 45, 45, 90 triangle and the hypotenuse is 46 miles. No I won’t use metric you dirty commie.

We *could* just straight up solve this with Pythagorean but for the sake of this post let’s grind out a trig function.

$$\sin(45^\circ)=\dfrac{x}{46\,\text{miles}}$$

$$\dfrac{\sqrt{2}}{2}=\dfrac{x}{46\,\text{miles}}$$

$$\dfrac{46\sqrt{2}\,\text{miles}}{2}=x$$

$$23\sqrt{2}\,\text{miles}=x$$

You can use your calculator for the rest but there you go. Since in 45° 45° 90° triangles the two sides are always equal, you now know all the sides of the triangle.

So that’s the unit circle. Just memorize it. It’s like 3 angles and 3 sets of coordinates. Don’t be a bitch.

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