Definition 2.2.1.
If the point \(P = \left(x,y \right) \) is specified by the angle \(\theta\) on the circumeference of a circle of radius \(r\text{,}\) then \(\cos (\theta ) = \frac{x}{r} \) and \(\sin (\theta ) = \frac{y}{r} \text{.}\)
Section 2.2 Generalized Definitions of Trigonometric Functions
We can generalize the definitions of the trigonometric functions so that they are applicable to circles of any size.
Notice that if we are on a unit circle, where \(r=1\text{,}\) then these definitions for \(\cos(\theta)\) and \(\sin(\theta)\) simplifly accordingly:
\begin{equation*}
\cos(\theta)=\frac{x}{r}=\frac{x}{1}=x
\end{equation*}
\begin{equation*}
\sin(\theta)=\frac{y}{r}=\frac{y}{1}=y
\end{equation*}
We can use Definition 2.2.1 to express the other four trigonometric functions in terms of \(x\text{,}\) \(y\text{,}\) and \(r\text{.}\)
\begin{align*}
\tan(\theta) \amp= \frac{\sin(\theta)}{\cos(\theta)} \\
\amp= \frac{y/r}{x/r} \\
\amp= \frac{y}{x}
\end{align*}
\begin{align*}
\csc(\theta) \amp= \frac{1}{\sin(\theta)} \\
\amp= \frac{1}{y/r} \\
\amp= \frac{r}{y}
\end{align*}
\begin{align*}
\sec(\theta) \amp= \frac{1}{\cos(\theta)} \\
\amp= \frac{1}{x/r} \\
\amp= \frac{r}{x}
\end{align*}
\begin{align*}
\cot(\theta) \amp= \frac{1}{\tan(\theta)} \\
\amp= \frac{1}{y/x} \\
\amp= \frac{x}{y}
\end{align*}
We summarize this in the following definition.
Definition 2.2.3.
If the point \(P = \left(x,y \right) \) is specified by the angle \(\theta\) on the circumeference of a circle of radius \(r\text{,}\) then we can define the six trigonemtric functions as follows.
\begin{equation*}
\cos (\theta ) = \frac{x}{r}
\end{equation*}
\begin{equation*}
\sin (\theta ) = \frac{y}{r}
\end{equation*}
\begin{equation*}
\tan (\theta ) = \frac{y}{x}
\end{equation*}
\begin{equation*}
\sec (\theta ) = \frac{r}{x}
\end{equation*}
\begin{equation*}
\csc (\theta ) = \frac{r}{y}
\end{equation*}
\begin{equation*}
\cot (\theta ) = \frac{x}{y}
\end{equation*}
Example 2.2.5.
Find the exact value of each of the six trigonometric functions of an angle \(\theta\) if \((-1,2)\) is a point on its terminal side.
Solution.
We need the values of \(x\text{,}\) \(y\text{,}\) and \(r\) to determine the exact value of each of the trigonemtric function. We are given \(x\) and \(y\text{,}\) but will need to find the value of \(r\text{.}\)
We can think of \(r\) as the hypotenuse of a right triangle whose horiztonal leg has a length of \(|-1|=1\) unit and vertical leg has a length of \(2\) units. We can use the Pythagorean Theorem to solve for \(r\text{:}\)
\begin{align*}
r^2 \amp= 1^2+2^2 \\
r^2 \amp= 5 \\
r \amp= \sqrt{5}
\end{align*}
Now that we have values for \(x\text{,}\) \(y\text{,}\) and \(r\text{,}\) we can use Definition 2.2.3 to state the exact values of each trigonmetric function.
\begin{align*}
\cos(\theta) \amp= \frac{x}{r} \\
\amp= \frac{-1}{\sqrt{5}} \\
\amp= \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\
\amp= -\frac{\sqrt{5}}{5}
\end{align*}
\begin{align*}
\sin(\theta) \amp= \frac{y}{r} \\
\amp= \frac{2}{\sqrt{5}} \\
\amp= \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} \\
\amp= \frac{2\sqrt{5}}{5}
\end{align*}
\begin{align*}
\tan(\theta) \amp= \frac{y}{x} \\
\amp= \frac{2}{-1} \\
\amp= -2
\end{align*}
\begin{align*}
\sec(\theta) \amp= \frac{r}{x} \\
\amp= \frac{\sqrt{5}}{-1} \\
\amp= -\sqrt{5}
\end{align*}
\begin{align*}
\csc(\theta) \amp= \frac{r}{y} \\
\amp= \frac{\sqrt{5}}{2}
\end{align*}
\begin{align*}
\cot(\theta) \amp= \frac{x}{y} \\
\amp= \frac{-1}{2} \\
\amp= -\frac{1}{2}
\end{align*}
We can use Definition 2.2.1 to express the coordinates of point \(P\) in Figure 2.2.2. We do this by solving the equations \(\cos (\theta ) = \frac{x}{r} \) and \(\sin (\theta ) = \frac{y}{r} \) for \(x\) and \(y\text{,}\) respectively:
\begin{equation*}
\cos(\theta)=\frac{x}{r} \implies x=r\cos(\theta)
\end{equation*}
\begin{equation*}
\sin(\theta)=\frac{y}{r} \implies y=r\sin(\theta)
\end{equation*}
We summarize this below.
Definition 2.2.7.
If the point \(P = \left(x,y \right) \) is specified by the angle \(\theta\) on the circumeference of a circle of radius \(r\text{,}\) then \(x=r \cos (\theta )\) and \(y=r \sin (\theta )\text{.}\)
Example 2.2.9.
Find the coordinates of the point on a circle with a radius of \(8\) units corresponding to an angle of \(\frac{7\pi}{4}\text{.}\)
Solution.
We are given the values of \(r\) and \(\theta\text{,}\) so we can use Definition 2.2.7 to determine the coordinate values.
Find \(x\text{:}\)
\begin{align*}
x \amp= r\cos\mathopen{}\left( \theta \right)\mathclose{} \\
\amp= 8\cos\mathopen{}\left( \frac{7\pi}{4}\right)\mathclose{} \\
\amp= 8 \cdot \frac{\sqrt{2}}{2} \\
\amp= 4\sqrt{2}
\end{align*}
Find \(y\text{:}\)
\begin{align*}
y \amp= r\sin\mathopen{}\left( \theta \right)\mathclose{} \\
\amp= 8\sin\mathopen{}\left( \frac{7\pi}{4}\right)\mathclose{} \\
\amp= 8 \mathopen{}\left( -\frac{\sqrt{2}}{2} \right)\mathclose{} \\
\amp= -4\sqrt{2}
\end{align*}
So the coordinates of the point on a circle with a radius of \(8\) units corresponding to an angle of \(\frac{7\pi}{4}\) are \(\mathopen{}\left( 4\sqrt{2}, -4\sqrt{2} \right)\mathclose{}\text{.}\)
Exercises Exercises
Six Trigonemtric Function Values.
In Exercises 1–2, find the exact value of each of the six trigonometric functions of an angle \(\theta\) if the given point is on its terminal side.
1.
\((3,4)\)
2.
\((-2,-6)\)
Find the coordinates.
In Exercises 3–4, find the coordinates of the point on a circle with the given radius corresponding to the given angle.
3.
\(r=3, \theta=\frac{7\pi}{6}\)
4.
\(r=10, \theta=\frac{\pi}{3}\)
