## Unit circle

Before learning about what a unit circle is, it helps to remember what is a number line. A number line is a straight endless line with origin and unitary length. ( OE – Unitary length, O – origin )

Now what would happen if we would wrap our endless line around a circle with radius 1?

Every point from the number line will end up on our circle. For every point on our number line, there is exactly one point on a circle. That circle is called a unit circle. The unit circle is a circle with a radius of 1. It is important that the radius of this circle is equal to 1.

As you know, you have positive and negative numbers on your number line. The positive numbers, (up from the origin in the picture) are replicated in a positive mathematical orientation (counterclockwise) and negative (downwards from the origin) are replicated in a negative mathematical orientation (clockwise).

Let’s remember what radians are. 1 radian is a part of a circle where length of an arc is equal to the radius. One whole circle has $ 2 \pi$ radians; one half of a circle has $\pi$ radians and so on.

Now that we remembered that, let’s look at our picture. You wrap an endless line around a circle. In some point you’ll start your second lap around it, and when you wrap it again, you’ll start third and so on in infinity. That means that infinitely many points from number line will fall into same places on a unit circle.

## Unit circle radians

If you have your number line marked with radians, this is how it would look:

First, you have a usual unit circle. In one quarter of a circle is $\frac{\pi}{2}$, in one half is $\pi$, in three quarters is $\frac{3 \pi}{2}$, and one whole is $2 \pi$.

Now what when you start another lap?

You’re again in zero, but now with 2π of the line around the circle. If you add another $\frac{\pi}{2}$ that will lead you into the point where the ‘old’ $\frac{\pi}{2}$ is, but now that value will be $2 \pi + \frac{\pi}{2} = \frac{5 \pi}{2}$. If you continue and add another $\frac{\pi}{2}$ you’ll find yourself in a point where the ‘old’ π lies. Now that point will be $\frac{5 \pi}{2} + \frac{\pi}{2} = 3 \pi$. And you continue like that.

This will lead us to:

In degrees you can conclude that $\frac{\pi}{2} = 90^{\circ}$, $\pi = 180^{\circ}$, $\frac{3 \pi}{2} = 270^{\circ}$, $ 2 \pi = 360^{\circ}$, $\frac{5 \pi}{2} = 450^{\circ}$ and so on.

By dividing radians into smaller and smaller parts we can determine measure of every angle.

Angles that are mostly used are 0, $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$ and so on.

Can you see a pattern here? If you only observe first and second quadrant you’ll notice that lines that are perpendicular on y-axis that go through $\frac{\pi}{3}$ and $\frac{2 \pi}{3}$ cut off equal parts of y-axis. The same applies with $\frac{3 \pi}{4}$ and $\frac{\pi}{4}$, and also with $\frac{5 \pi}{6}$ and $\frac{\pi}{6}$.

If you take a look at first and fourth quadrant you’ll notice that lines that are perpendicular of x-axis $\frac{\pi}{6}$ and $\frac{11 \pi}{6}$ cut off equal length of x – axis, and so on with other angles. This can help you in drawing them. For example, if you get a task to draw $\frac{5 \pi}{6}$, you can simply draw $\frac{\pi}{6}$ and translate it to second quadrant. Using this way you’ll only need to remember angles in first quadrant and translate them.

*Example 1*: Find following angles on the unit circle

When you have a fraction whose value is greater than two, that means that you’re starting another “lap” around the circle. When you are dealing with these kinds of values, you have to apply a process of finding the right measure of an angle. That means that you have to find an angle that suits given angle but in your first lap. You do that by subtracting with a multiple of 2π.

For example, let’s say you have $\frac{5 \pi}{2}$. $\frac{5 \pi}{2}$ is greater than $2 \pi$ by $\frac{\pi}{2}$. That means that you’ll finish first lap and end up in $\frac{\pi}{2}$.

*Example 2*: Find following angles on the unit circle

Since we have a minus in front of our values, we start looking from zero but in an opposite way. Whole circle is equal to $2 \pi$, which means that $ -\frac{\pi}{4}$ will have the same value as $ 2 \pi – \frac{\pi}{4} = \frac{7 \pi}{4}$, $- \pi$ as $ \pi$, and $ -2 \pi$ as 0.

## Trig unit circle

For every point on our unit circle we want to know the exact length from the origin to its projection on x and y axis.

### Sine and Cosine

The length from the origin to the projection of the point on the x axis is called the **cosine**, and the length from the origin to the projection of the point on y axis is called the **sine**.

Functions sine and cosine are functions whose domain is whole set of real numbers and codomain [-1, 1]. Their codomain is the set of real numbers between [-1, 1], because we’re dealing with points on unit circle whose radius is equal to 1.

How would you find an angle whose sine and cosine value you know? That is very simple. If you have a default sine value first you would find that value on y – axis and draw a parallel line from x – axis through that point. You’ll get two points on you unit line, which you’ll connect with the origin. Two angles whose one arm are those lines and second x – axis will be the angles you are looking for.

If you are given cosine value of an angle you’ll simply find that value on x – axis, draw a parallel line from y- axis through that point, connect the points you got on the unit circle and connect them with origin. Two angles whose one arm are those lines and second x – axis will be the angles you are looking for.

*Example 3*: Find angles whose

a) $ cos(x) = \frac{1}{2}$

b) $ sin(x) = – \frac{1}{2}$

c) $ sin(x) = 5$

d) $ cos(x) = 1$

*Solution:*

a)

b)

c) $ sin(x) = 5$. Where would you draw your five? If you’d follow these steps, they would lead you somewhere outside the unit circle. And no matter where you draw your parallels none of them would intersect unit circle. This is because of codomain of sine and cosine. Remember that their values can only be [-1,1].

d) This one is easy. $ cos(x) = 1$. This value can be found exactly on the number line, and is equal to 0.

### Tangent and Cotangent

Two other very important trigonometry functions are called **tangent** and **cotangent**.

They are derived functions, but still equally important to remember.

To get to the value of tangent, first we’ll draw a line parallel to y – axis that goes through point (1, 0).

To get to the value of cotangent, first we’ll draw a line parallel to x – axis that goes through point (0, 1).

Next, we draw a line from origin and the point on unit line whose value we’re looking for.

We’ll get two intersections: one with the line perpendicular to x – axis, and second with line perpendicular to y – axis.

The value of tangent in the point we are given is equal to the length from the intersection with the line perpendicular to x – axis and y – axis.

*Example 4: Find angles whose a) $ tan(x) = \frac{1}{6}$*

*b) $ cot(x) = \frac{1}{2}$*

*Solution:*

a) Generally, if $a$ is a constant, equation $tan(x)=a$ has infinitely many solutions. They are given with: $x=arctan(a)+k \pi, k \in \mathbf {Z}$, where $\arctan(a)={tan^{-1}(a)}$ is a value of arctangent function, the inverse function of restricted tangent function.

Now we see that the solution of given equation is: $x=\arctan(*\frac{1}{6})$ *+ $k \pi , k \in \mathbf{Z}$.

b) Generally, if $a$ is a constant, equation $cot(x)=a$ has infinitely many solutions. They are given with: $x=arccot(a)+k \pi, k \in \mathbf {Z}$, where $arccot(a)={cot^{-1}(a)}$ is a value of arccotangent function, the inverse function of restricted cotangent function.

Now we see that the solution of given equation is: $x=\arccot(*\frac{1}{2})$*+ $k \pi , k \in \mathbf{Z}$.

*Example 5: *Calculate the tangent of 60°.

*Solution:*

Since we know that $tan(x) = \frac {sin(x)}{cos(x)}$, we only have to know values of $sin(60°)$ and $cos(60°)$.

If we look at the unit circle, we can see that $sin(60°)= \frac{\sqrt{3}}{2}$ and $cos(60°)= \frac{1}{2}$.

Now we have:

$$tan(x) = \frac {sin(60°)}{cos(60°)} = \frac {\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$

$$=\frac{2 \cdot \sqrt{3}}{2}$$

$$=\sqrt{3}.$$

## Trigonometric functions of special angles

Special angles are angles that have relatively simple values. Following table is very important to remember. Consider Sine, Cosine, Tangent, and Cotangent:

This lesson may look a bit complicated to remember, but it really is not. You only need to remember first row. Cos goes opposite from sine, and tangent and cotangent are derived from them so you can always calculate them easily.