Section 5.2 Trigonometric Functions
Consider the point \(P\) = (\(x,y\)) where the angle of measure \(\theta\) intersects the unit circle. We use the coordinates \(x\) and \(y\) of this point to define six trigonometric functions of \(\theta\text{.}\)
We define the cosine of \(\theta\) to be the \(x\)-coordinate of this point and the sine of \(\theta\) to be the \(y\)-coordinate of this point. We use the abbreviations \(\cos\) for cosine and \(\sin\) for sine. Thus,
\begin{equation*}
\cos\theta= x \text{ and } \sin\theta=y.
\end{equation*}
Notice that the \(x\) and \(y\) coordinates of all points on the unit circle lie somewhere between \(-1\) and \(1\text{.}\) Thus,
\begin{equation*}
-1\leq \cos \theta\leq1 \text{ and } -1\leq \sin \theta\leq1
\end{equation*}
no matter what \(\theta\) is.
We can use geometry to determine \(\sin\left(\frac{\pi}{4}\right)\) and \(\cos\left(\frac{\pi}{4}\right)\text{.}\) Since \(\frac{\pi}{4}\) radians equals \(45^\circ\text{,}\) consdier a \(45^\circ-45^\circ-90^\circ\) triangle with a hypotenuse of length 1. Such a triangle must have legs each of length \(\frac{1}{\sqrt{2}}\text{.}\)
By moving this triangle into the unit circle and remembering that \(45^\circ\) equals \(\frac{\pi}{4}\) radians, we find that the corresponding point on the unit circle has \((x,y)\)-coordinates \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\text{.}\)
Thus
\begin{equation*}
\sin\frac{\pi}{4} = y = \frac{1}{\sqrt{2}} \text{ and } \cos\frac{\pi}{4}=x=\frac{1}{\sqrt{2}}\text{.}
\end{equation*}
We can find \(\sin\frac{\pi}{3}\text{,}\) \(\cos\frac{\pi}{3}\text{,}\) \(\sin\frac{\pi}{6}\text{,}\) and \(\cos\frac{\pi}{6}\) in a similar fashion by noting \(\frac{\pi}{3}\) radians equals \(60^\circ\) and \(\frac{\pi}{6}\) radians equals \(30^\circ\text{.}\) A \(30^\circ-60^\circ-90^\circ\) triangle with a hypotenuse of length 1 gives us the information we need.
Thus
\begin{equation*}
\sin\frac{\pi}{6}=\frac{1}{2} \text{ and } \cos\frac{\pi}{6}= \frac{\sqrt{3}}{2}\text{.}
\end{equation*}
and
\begin{equation*}
\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \text{ and } \cos\frac{\pi}{3}=\frac{1}{2}
\end{equation*}
In calculus, we measure angles in radians, and we often use the trig values we just found, so it will be helpful to memorize the “enhanced” unit circle in Figure 5.2.7
Four other trigonometric functions are defined using sine and consine. They are the secant (\(\sec\)), cosecant (\(\csc\)), tangent (\(\tan\)), and cotangent (\(\cot\)) trigonometric functions, defined as follows.
\begin{align*}
\sec\theta = \frac{1}{\cos\theta} \amp \text{ , } \amp \csc\theta = \frac{1}{\sin\theta}\\
\tan\theta = \frac{\sin\theta}{\cos\theta} \amp \text{ , } \amp \cot\theta = \frac{\cos\theta}{\sin\theta}
\end{align*}
Since \(\cos\theta\) and \(\sin\theta\) are \(0\) for some values of \(\theta\text{,}\) the trig functions \(\sec\theta\text{,}\) \(\csc\theta\text{,}\) \(\tan\theta\text{,}\) and \(\cot\theta\) are undefined for some values of \(\theta\text{.}\) For more information, see the graphs of the trig functions at the end of this section.
Find \(\cot\frac{\pi}{6}\text{.}\)
\begin{equation*}
\cot\frac{\pi}{6} = \frac{\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}\text{.}
\end{equation*}
The unit circle has four quadrants.
Note that a point in Quad II will have a negative \(x\)-value and a positive \(y\)-value. Thus, cosine is negative and sine is positive in Quad II. To remember which trig functions are positive in which quadrants, use the saying “All Students Take Calculus”.
Example 5.2.11. Finding a cosine.
Find \(\cos\frac{4\pi}{3}\text{.}\)
We will first find in what quadrant \(\frac{4\pi}{3}\) lies.
Due to symmetry, \(\frac{\pi}{3}\) and \(\frac{4\pi}{3}\) have the same coordinates except for the negative signs. Thus
\begin{equation*}
\cos\frac{4\pi}{3} = -\frac{1}{2}\text{.}
\end{equation*}
