Math Core

Secant and cosecant are the reciprocals of cosine and sine: $\sec x = \dfrac{1}{\cos x}$ and $\csc x = \dfrac{1}{\sin x}$ . You won't need a new table of values to graph them. If you can draw cosine and sine, you can draw these by "flipping" each height.

Graphing a reciprocal

Think about what taking a reciprocal does to a number between $-1$ and $1$ :

$1$ stays $1$ , and $-1$ stays $-1$ .
A number close to $0$ becomes a huge number with the same sign: $\dfrac{1}{0.1} = 10$ and $\dfrac{1}{-0.01} = -100$ .
$0$ has no reciprocal at all.
A fraction like $0.5$ becomes $2$ : smaller heights turn into larger ones.

Apply this to the cosine curve. Wherever $\cos x = 1$ (the peaks), $\sec x = 1$ too. As cosine drops toward $0$ , secant grows without bound. Where cosine equals $0$ , secant is undefined, and the graph has a vertical asymptote. So each hill of the cosine wave turns into a U-shaped branch opening up, and each valley turns into a branch opening down.

y = sec x drawn over its guide curve y = cos x (dashed), with asymptotes where cos x = 0.Open in grapher →

Notice that the two graphs touch exactly at the peaks and valleys of cosine, where the value is $\pm 1$ . Between the asymptotes, secant is always farther from the $x$ -axis than cosine.

The graph of y = sec x

Asymptotes where $\cos x = 0$ : $x = \dfrac{\pi}{2} + k\pi$
Local minimum points $(2k\pi, 1)$ on branches opening up; local maximum points $(\pi + 2k\pi, -1)$ on branches opening down
Range: $(-\infty, -1] \cup [1, \infty)$
Period: $2\pi$ (the same as cosine); even function
No zeros: $\dfrac{1}{\cos x}$ can never equal $0$

The graph of cosecant

Cosecant is built from sine in exactly the same way. The asymptotes fall where $\sin x = 0$ , at every multiple of $\pi$ . The branches touch the sine curve at its peaks and valleys: a branch opening up with its low point at $\left(\dfrac{\pi}{2}, 1\right)$ and a branch opening down with its high point at $\left(\dfrac{3\pi}{2}, -1\right)$ .

y = csc x over its guide curve y = sin x (dashed), with asymptotes at multiples of π.Open in grapher →

	$y = \sec x$	$y = \csc x$
Guide curve	$y = \cos x$	$y = \sin x$
Asymptotes	$x = \frac{\pi}{2} + k\pi$	$x = k\pi$
Range	$(-\infty, -1] \cup [1, \infty)$	$(-\infty, -1] \cup [1, \infty)$
Period	$2\pi$	$2\pi$
Symmetry	even	odd

Transformations: graph the guide curve first

To graph something like $y = 2\csc(2x)$ or $y = \sec\left(x - \dfrac{\pi}{3}\right) + 1$ , follow three steps:

Graph the matching sine or cosine curve, $y = A\sin\big(B(x - C)\big) + D$ or the cosine version, lightly or dashed.
Draw vertical asymptotes where that guide curve crosses its midline $y = D$ (that is where the sine or cosine part equals $0$ ).
At each peak and valley of the guide curve, draw a U-shaped branch that touches the guide there and bends away toward the neighboring asymptotes.

The period is $\dfrac{2\pi}{|B|}$ , just as for the guide curve. The branches now turn around at $D + |A|$ and $D - |A|$ , so the range becomes $(-\infty, D - |A|] \cup [D + |A|, \infty)$ .

Common mistake

Secant and cosecant have no amplitude and no maximum or minimum value: the branches go on forever. The number $A$ tells you where the branches turn around, not how high the graph goes. Also remember that asymptotes come from the zeros of the guide curve, not from its peaks.

Worked example: Evaluating from the reciprocal

Find $\sec \pi$ and $\csc\left(\dfrac{\pi}{6}\right)$ .

Solution. $\cos \pi = -1$ , so $\sec \pi = \dfrac{1}{-1} = -1$ . This is the high point of a downward branch.

$\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$ , so $\csc\left(\dfrac{\pi}{6}\right) = \dfrac{1}{1/2} = 2$ .

Worked example: A squeezed cosecant

Find the period, the asymptotes on $[0, \pi]$ , and the range of $y = 2\csc(2x)$ .

Solution. The guide curve is $y = 2\sin(2x)$ , with period $\dfrac{2\pi}{2} = \pi$ and amplitude $2$ .

Asymptotes where $\sin(2x) = 0$ : $2x = k\pi$ , so $x = \dfrac{k\pi}{2}$ . On $[0, \pi]$ these are $x = 0, \dfrac{\pi}{2}, \pi$ .
The guide peaks at $\left(\dfrac{\pi}{4}, 2\right)$ and bottoms out at $\left(\dfrac{3\pi}{4}, -2\right)$ , so the cosecant branches turn around there.
Range: $(-\infty, -2] \cup [2, \infty)$ .

Worked example: Shifted secant

Describe the branches of $y = \sec x - 2$ .

Solution. The guide curve is $y = \cos x - 2$ , which oscillates between $-1$ and $-3$ with midline $y = -2$ . Shifting down does not move the asymptotes, which stay at $x = \dfrac{\pi}{2} + k\pi$ . The branches opening up now have their low points at $(2k\pi, -1)$ , and the branches opening down have their high points at $(\pi + 2k\pi, -3)$ . The range is $(-\infty, -3] \cup [-1, \infty)$ .

Tip

To check whether a branch opens up or down, test one point near the turning point. For $y = -2\sec x$ at $x = 0$ : $y = -2 \cdot 1 = -2$ , and at $x = 0.5$ : $y = -2 \sec(0.5) \approx -2.28$ . The values get more negative, so that branch opens down.

Practice

Practice 1

What is the value of $y = \sec x$ at $x = \pi$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

List every $x$ in $[0, 2\pi]$ where $y = \csc x$ has a vertical asymptote. Separate answers with commas.

Separate answers with commas, e.g. 2, -5

Practice 3

What is the period of $y = \sec(4x)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

What is the range of $y = 3\csc x$ ?

Practice 5

Which function's graph has a branch opening upward whose lowest point is $\left(\dfrac{\pi}{2}, 1\right)$ ?

Practice 6

What is the smallest positive $x$ -value of a vertical asymptote of $y = \csc(2x)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

The graph of $y = \sec x - 2$ has branches opening upward. What is the $y$ -coordinate of the lowest point on one of those branches?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

For $y = -2\sec x$ , consider the branch between the asymptotes $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$ . It opens upward. What is the $y$ -coordinate of its lowest point?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.