##### The Graph of the Function \(y=\sec(t)\)

Recall that the secant function is the reciprocal of the cosine function. That is, \(\sec(t)=\frac{1}{\cos(t)}\text{.}\) Given that, it seems like a good place to begin our understanding of the graph of \(t=\sec(t)\) is to look at a graph of \(y=\cos(t)\text{.}\) Such a graph appears as FigureĀ 14.6.1.

The first thing we should note is that the period of the secant function will be the same as the period of the cosine function; that is, \(2\pi\text{.}\)

We should also note that \(\cos(t)=0\) at \(-\frac{5\pi}{2}\text{,}\) \(-\frac{3\pi}{2}\text{,}\) \(-\frac{\pi}{2}\text{,}\)\(\frac{\pi}{2}\text{,}\) \(\frac{3\pi}{2}\text{,}\) and \(\frac{5\pi}{2}\text{.}\) Consequently, at each of those values of \(t\text{,}\) the secant function has the form \(\frac{1}{0}\text{.}\) As is the case with rational functions (or any function, for that matter), at any value where the function returns a fraction of form \(\frac{\text{not zero}}{\text{zero}}\text{,}\) the graph of the function approaches a vertical asymptote. Consequently the graph of \(y=\sec(t)\) has a vertical asymptote everywhere \(\cos(t)=0\text{.}\) The vertical asymptote are, specifically: \(\,\,...,\,t=-\frac{5\pi}{2},\,t=-\frac{3\pi}{2},\,t=-\frac{\pi}{2},\,t=\frac{\pi}{2},\,t=\frac{3\pi}{2},\,t=\frac{5\pi}{2},\,...\text{.}\)

The vertical asymptotes that occur over \([-\frac{5\pi}{2},\frac{5\pi}{2}]\) have been added to the graph in FigureĀ 14.6.2.

At \(-2\pi\text{,}\) \(0\text{,}\) and \(2\pi\text{,}\) the value of the cosine function is \(1\text{,}\) so the value of the secant function is also \(1\text{.}\) Over the intervals \(\left(-\frac{5\pi}{2} ,-\frac{3\pi}{2}\right)\text{,}\) \(\left(-\frac{\pi}{2} ,\frac{\pi}{2}\right)\text{,}\) and \(\left(\frac{3\pi}{2} ,\frac{5\pi}{2}\right)\) the cosine function is always positive, so the secant function is also always positive. Consequently, at both ends of each of these intervals the secant curve will point upward along the asymptotes. These portions of the secant function have been added in FigureĀ 14.6.3.

At \(-\pi\) and \(\pi\text{,}\) the value of the cosine function is \(-1\text{,}\) so the value of the secant function is also \(-1\text{.}\) Over the intervals \(\left(-\frac{3\pi}{2} ,-\frac{\pi}{2}\right)\) and \(\left(\frac{\pi}{2} ,\frac{3\pi}{2}\right)\) the cosine function is always negative, so the secant function is also always negative. Consequently, at both ends of each of these intervals the secant curve will point downward along the asymptotes. The secant function has been completed in FigureĀ 14.6.4.

###### Example14.6.5

Sketch two complete periods of the function \(y=\frac{1}{2}\sec\left(3\left(t-\frac{\pi}{2}\right)\right)\text{.}\)

We begin by observing that \(\frac{1}{2}\sec\left(3\left(t-\frac{\pi}{2}\right)\right)=\frac{1}{2} \cdot \frac{1}{\cos\left(3\left(t-\frac{\pi}{2}\right)\right)}\text{.}\) We should also note that the \(y\)-coordinates at the local minimum points on the secant function will be \(\frac{1}{2}\) and the \(y\)-coordinates at the local maximum points will be \(-\frac{1}{2}\text{.}\) Consequently, if we want our curves to touch at the minimum and maximum points on the cosine function, the cosine function we should graph is \(y=\frac{1}{2}\cos\left(3\left(t-\frac{\pi}{2}\right)\right)\text{.}\)

The amplitude of the cosine function we will sketch is \(\frac{1}{2}\text{.}\) Since there is no vertical shift, the mid-line will be the \(x\)-axis and the minimum and maximum values of \(t\) will be, respectively, \(-\frac{1}{2}\) and \(\frac{1}{2}\text{.}\) We should note that we do not apply the word amplitude to the secant or cosecant function.

The period of both the cosine and the secant function we will sketch is \(\frac{2\pi}{3}\text{.}\) They both also have a rightward shift of \(\frac{\pi}{2}\) from their parent functions. Without the horizontal shift we would get two complete periods over the intervals \(\left(-\frac{2\pi}{3},0\right)\) and \(\left(0,\frac{2\pi}{3}\right)\text{.}\) With the shift we will two complete periods over the intervals \(\left(-\frac{2\pi}{3}+\frac{\pi}{2},0+\frac{\pi}{2}\right)\) and \(\left(0+\frac{\pi}{2},\frac{2\pi}{3}+\frac{\pi}{2}\right)\) which simplify to \(\left(-\frac{\pi}{6},\frac{\pi}{2}\right)\) and \(\left(\frac{\pi}{2},\frac{7\pi}{6}\right)\text{.}\)

One-quarter of the period is \(\frac{\pi}{6}\text{.}\) Beginning with \(-\frac{\pi}{6}\) and repeatedly adding \(\frac{\pi}{6}\) until we reach \(\frac{\pi}{2}\text{,}\) we see that the critical values of \(t\) over the interval \(\left(-\frac{\pi}{6},\frac{\pi}{2}\right)\) are \(-\frac{\pi}{6}\text{,}\) \(0\text{,}\) \(\frac{\pi}{6}\text{,}\) \(\frac{\pi}{3}\text{,}\) and \(\frac{\pi}{2}\text{.}\) Similarly, the critical values of \(t\) over the interval \(\left(\frac{\pi}{2},\frac{7\pi}{6}\right)\) are \(\frac{\pi}{2}\text{,}\) \(\frac{2\pi}{3}\text{,}\) \(\frac{5\pi}{6}\text{,}\) \(\pi\text{,}\) and \(\frac{7\pi}{6}\text{.}\)

Because we are graphing a positive-cosine function, the \(y\)-coordinates at the critical values of \(t\) will follow the patten max, mid-line, min, mid-line, max.

The cosine function is shown in FigureĀ 14.6.6.

The secant function and the cosine function will touch at the points \(\left(-\frac{\pi}{6},\frac{1}{2}\right)\text{,}\) \(\left(\frac{\pi}{6},-\frac{1}{2}\right)\text{,}\) \(\left(\frac{\pi}{2},\frac{1}{2}\right)\text{,}\) \(\left(\frac{5\pi}{6},-\frac{1}{2}\right)\text{,}\) and \(\left(\frac{7\pi}{6},\frac{1}{2}\right)\text{.}\)

The secant function will approach the vertical asymptotes \(t=0\text{,}\) \(t=\frac{\pi}{3}\text{,}\) \(t=\frac{2\pi}{3}\text{,}\) and \(t=\pi\text{.}\) Over the intervals \(\left(0,\frac{\pi}{3}\right)\) and \(\left(\frac{2\pi}{3},\pi\right)\) the secant curves will point downward along the asymptotes. Over \(\left(\frac{\pi}{3},\frac{2\pi}{3}\right)\) the secant curve will point upward along the asymptotes. Over the interval \(\left(-\frac{\pi}{6},0\right)\) the secant curve will point upward along the asymptote \(x=0\) and over the interval \(\left(\pi,\frac{7\pi}{6}\right)\) the secant curve will point upward along the asymptote \(t=\pi\text{.}\)

The secant function is graphed along with the cosine function in FigureĀ 14.6.7.