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
The unit circle is made up of infinite points.
Here is how to describe any point on the unit circle.
The technique is to "walk" to it: you need a starting point and an ending point.
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.
Here is an interactive example of a terminal point $P$ for positive $t$.
If $t$ were negative, we would travel clockwise.
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 \[t = 3\pi\qquad t = -\pi \qquad t=-\frac{\pi}{2}\]
Because $t = 3\pi = 2\pi + \pi$, we travel counterclockwise once around the unit circle, then halfway. The terminal point is $(-1, 0)$.
For $t=-\pi$, we travel clockwise halfway so the terminal point is $(-1, 0)$.
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:
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:
$\frac{7\pi}{6}$ overshoots $(-1, 0)$ by $\frac{\pi}{6}$, and $-\frac{5\pi}{6}$ is missing a distance of $\frac{\pi}{6}$ to reach $(-1, 0)$.
$\pm 2\pi$ 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!
Finding Terminal Points for any $t$
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
\[1. \ t = \frac{5\pi}{6} \qquad\qquad 2. \
t = \frac{7\pi}{4} \qquad\qquad 3. \
t = -\frac{2\pi}{3} \qquad\qquad 4.\
t = 5.80\]
$t$ is in the second quadrant. Therefore we drop down to the negative $x$-axis, or a half rotation $\pi$. Thus $\bar{t} = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
$t$ is in the fourth quadrant. Therefore we drop down to the positive $x$-axis, or a full rotation $2\pi$. Thus $\bar{t} = 2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.
$t$ 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 $\frac{\pi}{2}$.
$t$ is in the fourth quadrant. Therefore we drop down to the positive $x$-axis, or a full rotation $2\pi$.Thus $\bar{t} = 2\pi - 5.80 \approx 0.48$.
To find the terminal point $P$ for any value of $t \in \mathbb{R}$, do
Find the reference number $\bar{t}$.
Find the terminal point $Q(a,b)$ for $\bar{t}$.
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 \[1. \ t = \frac{5\pi}{6} \qquad\qquad 2. \
t = \frac{7\pi}{4} \qquad\qquad 3. \
t = -\frac{2\pi}{3}\]
$\bar{t} = \frac{\pi}{6}$
$Q$ is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$t$ lies in the second quadrant, with negative $x$, positive $y$. Thus $P$ is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\bar{t} = \frac{\pi}{4}$
$Q$ is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$
$t$ lies in the fourth quadrant, with positive $x$, negative $y$. Thus $P$ is $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$
$\bar{t} = \frac{\pi}{3}$
$Q$ is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$t$ 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: $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
Let's now define the trigonometric functions!
The Trigonometric Functions
Given the unit circle, for $t \in \mathbb{R}$ and it's associated terminal point $P(x, y)$, we can now define the trigonometric functions.
Let $t \in \mathbb{R}$ and let $P(x, y)$ be the corresponding terminal point on the unit circle. We define
\begin{align}
\sin t &= y & \cos t &= x & \tan t &= \frac{y}{x}, \ x \neq 0 \\
\csc t &= \frac{1}{y}, \ y \neq 0 & \sec t &= \frac{1}{x}, \ x \neq 0 & \cot t &= \frac{x}{y}, \ y \neq 0
\end{align}
The $\sin(t)$ function extracts the $y$ coordinate from $P(x, y)$! Look:
We will see more graphs in the next section!
Find the six trigonometric functions of the following: \[1. \ t = \frac{\pi}{3} \qquad\qquad 2. \ t = \frac{\pi}{2}\]
The corresponding terminal point is $P\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. From the definitions above, we have
\begin{align}
\sin\frac{\pi}{3} &= \frac{\sqrt{3}}{2} & \cos\frac{\pi}{3} &= \frac{1}{2} & \tan\frac{\pi}{3} &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \\
\csc \frac{\pi}{3} = \frac{2}{\sqrt{3}} &= \frac{2\sqrt{3}}{3} & \sec\frac{\pi}{3} &= 2 & \cot\frac{\pi}{3} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\end{align}
The corresponding terminal point is $P\left(0, 1\right)$. From the definitions above, we have
\begin{align}
\sin\frac{\pi}{2} &= 1 & \cos\frac{\pi}{2} &= 0 & \tan\frac{\pi}{2} &= \frac{1}{0} = \text{DNE} \\
\csc\frac{\pi}{2} &= \frac{1}{1} & \sec\frac{\pi}{2} = \frac{1}{0} &= \text{DNE} & \cot\frac{\pi}{2} &= \frac{0}{1} = 0
\end{align}
Now that we know what the trigonometric functions are, let's see some properties and how to use them.
The domains of the six trigonometric functions are
Function
Domain
$\sin, \cos$
$\mathbb{R}$
$\tan, \sec$
$\left\{t : t \in \mathbb{R} \text{ and } t \neq \frac{\pi}{2} + n\pi, n \in \mathbb{Z}\right\}$
$\cot, \csc$
$\left\{t : t \in \mathbb{R} \text{ and } t \neq n\pi, n \in \mathbb{Z}\right\}$
Why does $\tan(t)$ have a domain given above?
The domain is all real numbers except $t = \frac{\pi}{2} + n\pi$ where $n$ is an integer. First, consider $n = 0$. Then $t = \frac{\pi}{2} + 0 \pi$, which is only a quarter rotation counterclockwise. Rotating a quarter gives $P(0, 1)$, which has $x = 0$. Since $\tan$ divides by $x$, you cannot divide by zero, implying $t = \frac{\pi}{2}$ is not in the domain.
When $n = 1$, $t = \frac{\pi}{2} + \pi$, which is our quarter rotation as before, but then another half circle rotation. This again forces the $x$ value to be zero. So we can see that in general, the points where $x = 0$ are where the domain are excluded!
Values of the Trigonometric Functions
What are the signs of the trigonometric functions in each quadrant?
Given an arbitrary $t \in \mathbb{R}$, how can we quickly find, for example, $\tan(t)$? This is where the reference number + the sign of the function comes in handy. Below is the three step process:
If $f(t)$ is an arbitrary trigonometric function, $t \in \mathbb{R}$, we find $f(t)$ in three steps:
Find the reference number $\bar{t}$.
Determine the sign of the trigonometric function based on the quadrant.
$f(\bar{t})$ has the same value as $f(t)$, except for sign. Correct the sign using the correct sign in Step 2.
Find each value: \[\cos\left(\frac{2\pi}{3}\right) \qquad\qquad \tan\left(-\frac{\pi}{3}\right) \qquad\qquad \sin\left(\frac{19\pi}{4}\right)\]
Fundamental Trigonometric Identities
Reciprocal identities:
\[\csc t = \frac{1}{\sin t} \qquad\qquad \sec t = \frac{1}{\cos t} \qquad\qquad \cot t = \frac{1}{\tan t} \qquad\qquad \tan t = \frac{\sin t}{\cos t} \qquad\qquad \cot t = \frac{\cos t}{\sin t}\]
Pythagorean identities:
\[\sin^2 t + \cos^2 t = 1 \qquad\qquad \tan^2t + 1 = \sec^2 t \qquad\qquad 1 + \cot^2 t = \csc^2 t\]
Key ideas:
You only need to memorize $\sin, \cos, \tan$. The other three trigonometric functions are just reciprocals of these three.
You can derive the Pythagorean identities from the unit circle.
The notation $\sin^2t$ means $(\sin t)^2$. The entire quantity of $\sin$ must be squared.
If you are given one trigonometric function's value, you can find the rest.
Derive the three Pythagorean identities, starting with $\sin^2 t + \cos^2 t= 1$.