This page explains how to sketch the secant and cosecant functions of the form \[ y = a \sec\left(k(x - d)\right) \quad \text{and} \quad y = a \csc\left(k(x - d)\right) \] Detailed examples are provided to illustrate their graphs, transformations, and key characteristics step by step.
Range: \[ (-\infty , -1] \cup [1 , +\infty) \]
Period: \[ 2\pi \]
Vertical asymptotes of \( y = \sec(x) = \frac{1}{\cos(x)} \) occur at the zeros of \( \cos(x) \), given by: \[ x = \frac{\pi}{2} + k\pi,\quad k = 0, \pm1, \pm2, \dots \]
Vertical asymptotes of \( y = \csc(x) = \frac{1}{\sin(x)} \) occur at the zeros of \( \sin(x) \), given by: \[ x = k\pi,\quad k = 0, \pm1, \pm2, \dots \]
To sketch basic secant and cosecant functions, use the identities: \[ y = \sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad y = \csc(x) = \frac{1}{\sin(x)} \] These identities help identify the locations of vertical asymptotes.
All zeros of \( \cos(x) \) (which appear in the denominator) are vertical asymptotes of \( \sec(x) \).
All zeros of \( \sin(x) \) (which appear in the denominator) are vertical asymptotes of \( \csc(x) \).
Sketch the graph of \( y = \sec(2x - \pi/3) \) over one period.
Graphing Parameters
Range: \( (-\infty , -1] \cup [ 1, +\infty) \)
Period: \[ \frac{2\pi}{2} = \pi \]
Vertical asymptotes are given by solving the equation: \[ 2x - \frac{\pi}{3} = \frac{\pi}{2} + k\pi \] which gives: \[ x = \frac{5\pi}{12} + \frac{k\pi}{2}, \quad k = 0 , \pm1, \pm2, \ldots \]
Horizontal Shift: Because of the term \( -\pi/3 \), the graph is shifted horizontally. We first rewrite the given function as: \[ y = \sec\left[2\left(x - \frac{\pi}{6}\right)\right] \] We can now identify the horizontal shift as \( \frac{\pi}{6} \) to the right.
We sketch \( y = \sec(2x - \pi/3) \) by translating the graph of \( y = \sec(2x) \) by \( \pi/6 \) to the right (red graph below), so that the sketched period starts at \( \pi/6 \) and ends at \( \pi/6 + \pi = 7\pi/6 \), which is one full period equal to \( \pi \).
Sketch the graph of \[ y = -3 \csc\left(\frac{x}{2} + \frac{\pi}{2}\right) \] over one period.
Graphing Parameters
Range: \[ (-\infty , -3] \cup [ 3, +\infty) \]
Period: \[ \frac{2\pi}{|k|} = \frac{2\pi}{\frac{1}{2}} = 4\pi \]
Vertical asymptotes are given by solving: \[ \frac{x}{2} + \frac{\pi}{2} = k\pi \] Solving for \(x\), we get: \[ x = (2k - 1)\pi, \quad k = 0, \pm 1, \pm 2, \ldots \]
Horizontal Shift: Because of the term \(\frac{\pi}{2}\), the graph is shifted horizontally. Rewriting the function: \[ y = -3 \csc\left(\frac{1}{2}(x + \pi)\right) \] This shows a horizontal shift of \(\pi\) units to the left.
We sketch
\[
y = -3 \csc\left(\frac{x}{2} + \frac{\pi}{2}\right)
\]
by translating the graph of
\[
y = -3 \csc\left(\frac{x}{2}\right)
\]
to the left by \(\pi\), so that the sketched period starts at \(-\pi\) and ends at \(\pi + 4\pi = 3\pi\), which corresponds to one full period.
