5.1: The Unit Circle

The Unit Circle

Let's begin our foray into the world of trigonometric functions! Before we define the trigonometric functions, we need to discuss a tool to help us define them: the unit circle.

The unit circle is the circle of radius 1 centered at the origin in the $xy$-plane. The equation is $x^2 + y^2 = 1$
Is the point $P\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{6}}{3}\right)$ on the unit circle?
A point is on the unit circle if it satisfies the unit circle equation $x^2 + y^2 = 1$. We have $\left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{6}}{3}\right)^2 = \frac{3}{9} + \frac{6}{9} = 1$ so this point is on the unit circle.

Terminal Points and the Reference Number

Let $t$ be a real number. If $t \geq 0$, travel $t$ units along the unit circle counterclockwise from $(1, 0)$ and mark the destination point $P$. If $t < 0$, travel $\lvert t \rvert$ units along the unit circle clockwise from $(1, 0)$ and mark the destination point $P$.
The point $P$ is called the terminal point.
The $t$ value is sometimes called radian measure.
Here is an interactive example of a terminal point $P$ for positive $t$.
If $t$ were negative, we would travel clockwise.

Goal: Determine terminal points for any t value on the unit circle.

• The circumference of the unit circle is $C = 2\pi r = 2\pi$. This is why we saw $t = \pi$ when we traversed half of the circle above.
• This means if $t = \frac{\pi}{2}$, we would have traveled a quarter counterclockwise.

Find the terminal points for $t = 3\pi\qquad t = -\pi \qquad t=-\frac{\pi}{2}$
1. Because $t = 3\pi = 2\pi + \pi$, we travel counterclockwise once around the unit circle, then halfway. The terminal point is $(-1, 0)$.
2. For $t=-\pi$, we travel clockwise halfway so the terminal point is $(-1, 0)$.
3. For $t=-\frac{\pi}{2}$, we travel clockwise a quarter around the unit circle; the terminal point is $(0, -1)$.

Important! The terminal points associated with $t = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$ are common values you should commit to memory. Here they are:

$t$ Terminal point
$0$ $(1, 0)$
$\frac{\pi}{6}$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\frac{\pi}{4}$ $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$
$\frac{\pi}{3}$ $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$\frac{\pi}{2}$ $(0, 1)$

Show the terminal point for $t = \frac{\pi}{4}$ is the one stated above.
Let's derive the terminal point $P$ for $t = \frac{\pi}{4}$. We know the terminal point lies halfway between $(0, 1)$ and $(1, 0)$. The unit circle is symmetric around the line $y = x$ so this line must cut through the terminal point $P$.
We know $x^2 + y^2 = 1$, so substitute $y = x$ into the equation and solve for $x:$ \begin{align} x^2 + y^2 &= 1\\ x^2 + x^2 &= 1\\ 2x^2 &= 1\\ x^2 &= \frac{1}{2}\\ x &= \pm \frac{1}{\sqrt{2}} \end{align} When $t = \frac{\pi}{4}$, $P$ is in the first quadrant so $x$ must be $\frac{1}{\sqrt{2}}$. Since $y = x$ we know $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = P\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ after rationalizing the denominator. Similar arguments can be used to derive the terminal points for $t = \frac{\pi}{6}, \frac{\pi}{3}$.
Given $t = \frac{7\pi}{6}$ and its terminal point $P$, state two negative $t$ values that correspond to $P$.
One solution is $t = -\frac{5\pi}{6}$. Two ways to look at it:
1. $\frac{7\pi}{6}$ overshoots $(-1, 0)$ by $\frac{\pi}{6}$, and $-\frac{5\pi}{6}$ is missing a distance of $\frac{\pi}{6}$ to reach $(-1, 0)$.
2. $\pm 2\pi$ is a full rotation around the unit circle. Since $\frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6}$, this means $-\frac{5\pi}{6}$ also shares $P$ as its terminal point.

In general, all solutions are $t = \frac{7\pi}{6} + 2k\pi$ where $k$ is an integer.

The terminal points in the first quadrant are all you need to derive the rest of the terminal points in the rest of the quadrants! See below for the first technique.

Find the terminal points associated with $t = -\frac{\pi}{4} \qquad t = \frac{3\pi}{4} \qquad t = -\frac{5\pi}{6}$
• From above we know that when $t = \frac{\pi}{4}, P\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ is the terminal point. Since $t = -\frac{\pi}{4}$ is rotating counterclockwise, the terminal point $Q$ will share the same $x$-coordinate, but have a negative $y$-value instead. Thus the terminal point is $Q\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$.
• Let $P$ be the terminal point for $t = \frac{\pi}{4}$ and $Q$ be our desired terminal point.
Note that $t = \frac{3\pi}{4} = \frac{\pi}{2} + \frac{\pi}{4}$, meaning we first travel a quarter counterclockwise. Then by traveling a distance of $\frac{\pi}{4}$, we know that $Q$ has the same coordinates as $P$, except for sign due to being in a different quadrant. Clearly, the $y$-coordinate must be the same, but the $x$-coordinate is negative so $Q\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.
• Let $Q$ be our desired terminal point. From the previous example, we know $t = \frac{7\pi}{6}$ also has $Q$ as its terminal point. Note that $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, meaning rotate counterclockwise halfway, then $\frac{\pi}{6}$. As a result, $Q$ must have the same terminal point as $t=\frac{\pi}{6}$, except for sign. Since both $x$ and $y$ are negative, we know $Q\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.

The previous example is particularly insightful: if we can decompose $t$ into quarter/half rotations and perhaps some rotation $0 \leq q \leq \frac{\pi}{4}$, we can use symmetry arguments to find the terminal point. This is the exact reason why the table above is all you need to derive terminal points. Let's discuss how to do this.

Let $t$ be a real number and $P$ be the associated terminal point. The reference number $\bar{t}$ associated with $t$ is the shortest distance from $P$ to the $x$-axis.
Here is an interactive example of the reference number for $0 \leq t \leq 2\pi$.
If $t$ were negative, we would travel clockwise.
Find the reference numbers for
$1. \ t = \frac{5\pi}{6} \qquad\qquad 2. \ t = \frac{7\pi}{4} \qquad\qquad 3. \ t = -\frac{2\pi}{3} \qquad\qquad 4.\ t = 5.80$

1. $t$ is in the second quadrant. Therefore we drop down to the negative $x$-axis, or a half rotation $\pi$. Thus $\bar{t} = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
2. $t$ is in the fourth quadrant. Therefore we drop down to the positive $x$-axis, or a full rotation $2\pi$. Thus $\bar{t} = 2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.
3. $t$ is in the third quadrant. Therefore we drop down to the negative $x$-axis, or a half rotation $\pi$. Thus $\bar{t} = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$.
• Note: even though $t$ is negative, we are only interested in positive distance for $\bar{t}$. Make sure the $\bar{t}$ value lies between $0$ and $\frac{\pi}{2}$.
4. $t$ is in the fourth quadrant. Therefore we drop down to the positive $x$-axis, or a full rotation $2\pi$.Thus $\bar{t} = 2\pi - 5.80 \approx 0.48$.

The following method puts the technique we used three examples ago into a straightforward way to find the reference number for any terminal point.

To find the terminal point $P$ for any value of $t \in \mathbb{R}$, do
1. Find the reference number $\bar{t}$.
2. Find the terminal point $Q(a,b)$ for $\bar{t}$.
3. The terminal point for $t$ has form $P(\pm a, \pm b)$, depending on which quadrant you are in. Fix the sign according to the quadrant.
Find the terminal point for $1. \ t = \frac{5\pi}{6} \qquad\qquad 2. \ t = \frac{7\pi}{4} \qquad\qquad 3. \ t = -\frac{2\pi}{3}$
1. $\bar{t} = \frac{\pi}{6}$
2. $Q$ is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
3. $t$ lies in the second quadrant, with negative $x$, positive $y$. Thus $P$ is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
1. $\bar{t} = \frac{\pi}{4}$
2. $Q$ is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$
3. $t$ lies in the fourth quadrant, with positive $x$, negative $y$. Thus $P$ is $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$
1. $\bar{t} = \frac{\pi}{3}$
2. $Q$ is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
3. $t$ lies in the third quadrant, with negative $x$, negative $y$. Thus $P$ is $\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right))$
Find the terminal point for $t = \frac{65\pi}{6}.$
If $t < -2\pi$ or $t > 2\pi$, bring $t$ closer to zero by subtracting or adding $2\pi$. This does not change the terminal point for $t$ because these are full rotations around the unit circle.
For this case, the decomposition is $t = \frac{65\pi}{6} = \frac{60 \pi}{6} + \frac{5 \pi}{6} = 10\pi + \frac{5 \pi}{6} = 5(2\pi) + \frac{5 \pi}{6}$ The five initial rotations around the unit circle can be neglected. Thus $t$ has the same terminal point as #1 in the previous example: $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.