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: \[\cos\left(\frac{2\pi}{3}\right) \qquad\qquad \tan\left(-\frac{\pi}{3}\right) \qquad\qquad \sin\left(\frac{19\pi}{4}\right)\]
Even-Odd Properties
$\sin, \csc, \tan, \cot$ are odd functions. $\cos, \sec$ are even:
\begin{array}{ccc}
\sin(-t) = -\sin t & \cos(-t) = \cos t & \tan(-t) = -\tan t\\
\csc(-t) = -\csc t & \sec(-t) = \sec t & \cot(-t) = -\cot t\\
\end{array}
Use the unit circle to show why $\sin$ is odd and $\cos$ is even.
Use the even-odd properties to find \[\sin\left(-\frac{\pi}{6}\right) \qquad \qquad \cos\left(-\frac{\pi}{4}\right)\]
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 $\sin^2 t + \cos^2 t= 1$.