Section 9.1 Trigonometric Functions
In
Section 8.2 we defined the six trigonometric function values of an
acute angle
\(0 \lt \alpha \lt 90^\circ\text{.}\) In this section we extend our definition of these functions to
any angle
\(\theta\text{.}\) Much of this is summarized in the video below.
Suppose \(\theta\) is any angle drawn in standard position in the \(xy\)-plane and suppose \((x,y) \neq (0,0)\) is any point on the terminal ray of \(\theta\text{.}\) Let
\begin{equation*}
r = \sqrt{x^2+y^2} \gt 0
\end{equation*}
denote the radial distance form the origin
\((0,0)\) to the point
\((x,y)\text{.}\) Then
\((x,y)\) is a point on the circle
\(x^2+y^2=r^2\text{.}\) This setup is summarized in the figure below. Spend some time reviewing it before proceeding.
Definition 9.1. The Six Trigonometric Functions of an Angle.
We define sine, cosine, and tangent along with their respective reciprocals:
\begin{gather*}
\sin(\theta) = \frac{y}{r}\\
\cos(\theta) = \frac{x}{r}\\
\tan(\theta) = \frac{y}{x}
\end{gather*}
\begin{gather*}
\csc(\theta) = \frac{r}{y}\\
\sec(\theta) = \frac{r}{x}\\
\cot(\theta) = \frac{x}{y}
\end{gather*}
Note that tangent and secant are defined only when
\(x\neq 0\text{,}\) while cosecant and cotangent require
\(y \neq 0\text{.}\)
We maintain the familar reciprocal identities:
\begin{equation}
\csc(\theta) = \frac{1}{\sin(\theta)}, \quad \sec(\theta) = \frac{1}{\cos(\theta)}, \quad \cot(\theta) = \frac{1}{\tan(\theta)}\tag{9.1}
\end{equation}
and ratio identities:
\begin{equation}
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}.\tag{9.2}
\end{equation}
Example 9.2. Finding Trigonometric Function Values.
Suppose the terminal side of an angle \(\theta\) in standard position includes the point \((-3,2)\text{.}\) Let’s find the six trigonometric function values of \(\theta\text{.}\)
Start by drawing a picture using the given point.
We see that
\(\theta\) is a Quadrant II angle. We have
\(x=-3\) and
\(y=2\text{,}\) so that
\begin{equation*}
r = \sqrt{(-3)^2+2^2} = \sqrt{9+4}=\sqrt{13}.
\end{equation*}
Then
\begin{align*}
&\sin(\theta) = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}\\
&\cos(\theta) = \frac{-3}{\sqrt{13}} = -\frac{3\sqrt{13}}{13}\\
&\tan(\theta) = \frac{2}{-3} = -\frac{2}{3}
\end{align*}
\begin{align*}
&\csc(\theta) = \frac{\sqrt{13}}{2}\\
&\sec(\theta) = - \frac{\sqrt{13}}{3}\\
&\cot(\theta) = - \frac{3}{2}
\end{align*}
It’s tempting to think we can measure the angle
\(\theta\) in the previous example given the point
\((-3,2)\text{.}\) But keep in mind there are infinitely many coterminal angles that share that point on its terminal side.
Any of the above angles could be the angle in the example. This does not affect the resulting trigonometric function values:
coterminal angles have the same trigonometric function values.
Principle 9.3. Trigonometric Function Values of Coterminal Angles.
Suppose \(\alpha\) and \(\beta\) are coterminal angles. Then, all trigonometric function values of \(\alpha\) and \(\beta\) are equal, when they are defined.
Example 9.4. Finding Trigonometric Function Values.
The terminal ray of a Quadrant IV angle \(\theta\) in standard position is parallel to the line \(y = -x/2\text{.}\) Find the sine, cosine, and tangent of \(\theta\text{.}\)
Solution.
Draw a picture of the line
\(y=-x/2\) and the terminal ray in the correct quadrant.
Pick
any point on the terminal ray. For instance, if I take
\(x=2\text{,}\) then
\(y=-2/2 = -1\) so that
\((2,-1)\) is a Quadrant IV point on the terminal ray. Then
\begin{equation*}
r = \sqrt{2^2+(-1)^2} = \sqrt{4+1} = \sqrt{5}.
\end{equation*}
We have
\begin{align*}
&\sin(\theta) = \frac{-1}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\
&\cos(\theta) = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\\
&\tan(\theta) = \frac{-1}{2} = - \frac{1}{2}.
\end{align*}
Trigonometric function values of quadrantal angles are easy to find, but some may be undefined.
Example 9.5. Finding Trigonometric Function Values of a Quadrantal Angle.
Find the trigonometric function values of \(\theta = -5 \pi/2\text{,}\) if defined.
Solution.
We are given the angle measure and can plot this directly. Recall that
\(\pi/2\) is one quarter of a revolution. We should count
\(5\) of these quarters in the clockwise orientation.
Pick any point on the terminal side. For instance
\((0,-1)\) will suffice. Then
\(r = 1\text{.}\)
\begin{align*}
&\sin(\theta) = \frac{-1}{1} = -1\\
&\cos(\theta) = \frac{0}{1} = 0\\
&\tan(\theta) = {\color{red}\frac{-1}{0}}, \text{undefined}\\
&\csc(\theta) = \frac{1}{-1} =-1\\
&\sec(\theta) = {\color{red}\frac{1}{0}}, \text{undefined}\\
&\cot(\theta) = \frac{0}{-1} = 0
\end{align*}
