3.11 - The Secant, Cosecant, and Cotangent Functions

Overview

Sine, cosine, and tangent each have a reciprocal function: secant, cosecant, and cotangent. Each one is 1 divided by a function you already know, so every feature of its graph (where it's undefined, its range, where it increases or decreases) comes straight from the original function. On the AP exam you need to evaluate these functions, connect their graphs to sine, cosine, and tangent, and describe their key characteristics.

The three reciprocal functions

The reciprocal of a number is 1 divided by that number. The reciprocal of 4 is

\frac{1}{4}

, and the reciprocal of

-\frac{2}{3}

-\frac{3}{2}

. The three new trig functions are defined the same way.

Function	Definition	Undefined when
Secant: $\sec\theta$	$\dfrac{1}{\cos\theta}$	$\cos\theta = 0$
Cosecant: $\csc\theta$	$\dfrac{1}{\sin\theta}$	$\sin\theta = 0$
Cotangent: $\cot\theta$	$\dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$	$\sin\theta = 0$

To keep the pairs straight, notice that each pair has exactly one “co”: secant goes with cosine, cosecant goes with sine, and cotangent goes with tangent.

Values from the unit circle

If the terminal ray of

\theta

meets the unit circle at

(x, y)

, then

\cos\theta = x

and

\sin\theta = y

. So:

\sec\theta = \frac{1}{x} \qquad \csc\theta = \frac{1}{y} \qquad \cot\theta = \frac{x}{y}

A reciprocal always has the same sign as the original value. Cosine is negative in Quadrants II and III, so secant is negative there too. Sine is negative in Quadrants III and IV, so cosecant is too. Cotangent has the same sign as tangent: positive in Quadrants I and III, negative in Quadrants II and IV.

Example 1: Exact values

Find the exact value of each expression, or state that it is undefined.

$\sec\frac{\pi}{3}$ : since $\cos\frac{\pi}{3} = \frac{1}{2}$ , $\sec\frac{\pi}{3} = \dfrac{1}{1/2} = 2$ .

$\csc\frac{5\pi}{4}$ : since $\sin\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$ , $\csc\frac{5\pi}{4} = -\dfrac{2}{\sqrt{2}} = -\sqrt{2}$ .

$\cot\frac{5\pi}{6}$ : $\cot\frac{5\pi}{6} = \dfrac{\cos(5\pi/6)}{\sin(5\pi/6)} = \dfrac{-\sqrt{3}/2}{1/2} = -\sqrt{3}$ .

$\cot\frac{3\pi}{2}$ : $\cos\frac{3\pi}{2} = 0$ and $\sin\frac{3\pi}{2} = -1$ , so $\cot\frac{3\pi}{2} = \dfrac{0}{-1} = 0$ .

$\csc\pi$ : $\sin\pi = 0$ , so $\csc\pi$ is undefined.

Part 4 is a common trap.

\tan\frac{3\pi}{2}

is undefined, but

\cot\frac{3\pi}{2}

is defined and equals 0. Always compute cotangent as

\frac{\cos\theta}{\sin\theta}

, not as “1 over tangent.”

The secant function

f(\theta) = \sec\theta

is defined everywhere except where

\cos\theta = 0

, which happens at

\theta = \frac{\pi}{2} + k\pi

for any integer

k

. The graph shows

y = \sec\theta

in black, with

y = \cos\theta

in gray for comparison.

3.11 - The Secant, Cosecant, and Cotangent Functions

Vertical asymptotes where cosine is 0. As $\cos\theta$ gets close to 0, its reciprocal gets huge in size, so the graph shoots up or down along the dashed lines $\theta = \frac{\pi}{2} + k\pi$ .

Range: $(-\infty, -1] \cup [1, \infty)$ . Cosine is always between $-1$ and 1, and 1 divided by a nonzero number between $-1$ and 1 is at least 1 in size. So secant never takes a value strictly between $-1$ and 1.

Where the graphs touch. Where $\cos\theta = 1$ or $-1$ , secant has the same value, since $\frac{1}{1} = 1$ and $\frac{1}{-1} = -1$ . These are the dots: the bottoms of the upward U shapes and the tops of the downward ones.

Period and symmetry. Secant has period $2\pi$ , like cosine. It is an even function, $\sec(-\theta) = \sec\theta$ , because cosine is even, so its graph is symmetric about the y-axis.

Increasing and decreasing flip. Wherever cosine is increasing, secant is decreasing, and the other way around. That's because $y = \frac{1}{x}$ is decreasing on each side of $x = 0$ : a bigger input gives a smaller output, as long as you don't cross 0.

The cosecant function

f(\theta) = \csc\theta

is defined everywhere except where

\sin\theta = 0

, at

\theta = k\pi

. That includes

\theta = 0

, so the y-axis itself is one of the asymptotes.

Vertical asymptotes at $\theta = k\pi$ , where sine is 0.

Range: $(-\infty, -1] \cup [1, \infty)$ , for the same reason as secant.

Touch points at $\left(\frac{\pi}{2}, 1\right)$ and $\left(\frac{3\pi}{2}, -1\right)$ , where sine is 1 or $-1$ .

Period and symmetry. Cosecant has period $2\pi$ . It is an odd function, $\csc(-\theta) = -\csc\theta$ , because sine is odd.

Same shape as secant, shifted. The sine graph is the cosine graph shifted right by $\frac{\pi}{2}$ , so the cosecant graph is the secant graph shifted right by $\frac{\pi}{2}$ .

Example 2: Describing secant on an interval

Describe

f(\theta) = \sec\theta

on the interval

-\frac{\pi}{2} < \theta < \frac{3\pi}{2}

: where it is undefined, where it is increasing or decreasing, and its relative extrema.

Step 1: Find where it's undefined.

\cos\theta = 0

\theta = \frac{\pi}{2}

inside this interval, so

f

has a vertical asymptote there.

Step 2: Use the behavior of cosine, flipping each direction:

Interval	$\cos\theta$ is	$\sec\theta$ is
$-\frac{\pi}{2} < \theta < 0$	positive, increasing	positive, decreasing
$0 < \theta < \frac{\pi}{2}$	positive, decreasing	positive, increasing
$\frac{\pi}{2} < \theta < \pi$	negative, decreasing	negative, increasing
$\pi < \theta < \frac{3\pi}{2}$	negative, increasing	negative, decreasing

Step 3: Find the extrema. At

\theta = 0

f

switches from decreasing to increasing, so it has a relative minimum of

\sec 0 = 1

. At

\theta = \pi

f

switches from increasing to decreasing, so it has a relative maximum of

\sec\pi = -1

Answer:

f

is decreasing on

\left(-\frac{\pi}{2}, 0\right)

and

\left(\pi, \frac{3\pi}{2}\right)

, and increasing on

\left(0, \frac{\pi}{2}\right)

and

\left(\frac{\pi}{2}, \pi\right)

. It has a relative minimum of 1 at

\theta = 0

and a relative maximum of

-1

\theta = \pi

. The relative maximum value is smaller than the relative minimum value, which is possible because they sit on separate pieces of the graph.

Don't combine the two increasing intervals into

(0, \pi)

. The function isn't defined at

\frac{\pi}{2}

, so it can't be increasing across it.

The cotangent function

Cotangent is the reciprocal of tangent:

\cot\theta = \frac{1}{\tan\theta}

where

\tan\theta \neq 0

. Since

\tan\theta = \frac{\sin\theta}{\cos\theta}

, flipping it gives the equivalent form

\cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \sin\theta \neq 0

Use this second form. It also covers angles like

\theta = \frac{\pi}{2}

, where tangent is undefined but

\cot\frac{\pi}{2} = \frac{0}{1} = 0

Vertical asymptotes where $\tan\theta = 0$ , which is where $\sin\theta = 0$ : at $\theta = k\pi$ , including the y-axis.

Zeros where $\cos\theta = 0$ : at $\theta = \frac{\pi}{2} + k\pi$ . These are exactly where tangent has its asymptotes.

Decreasing between consecutive asymptotes. On $(0, \pi)$ the graph falls from $+\infty$ down to $-\infty$ , and the same thing repeats between every pair of neighboring asymptotes. Tangent is increasing between its asymptotes, and taking the reciprocal flips that.

Range and period. The range is all real numbers, and the period is $\pi$ , the same as tangent. Cotangent is an odd function.

Example 3: Behavior near an asymptote

Describe the behavior of

h(\theta) = \cot\theta

\theta

approaches

\pi

from each side.

From the left (

\theta

slightly less than

\pi

, in Quadrant II):

\cos\theta

is close to

-1

and

\sin\theta

is a tiny positive number. A number near

-1

divided by a tiny positive number is a huge negative number.

\lim_{\theta \to \pi^-} \cot\theta = -\infty

From the right (

\theta

slightly more than

\pi

, in Quadrant III):

\cos\theta

is still close to

-1

, but

\sin\theta

is a tiny negative number, so the quotient is a huge positive number.

\lim_{\theta \to \pi^+} \cot\theta = \infty

This matches the graph: the branch on

(0, \pi)

falls toward

-\infty

, and the next branch comes down from

+\infty

Example 4: Using one value to find the others (AP-style)

\cot\theta = -\frac{4}{3}

and

\frac{\pi}{2} < \theta < \pi

, find

\sec\theta

and

\csc\theta

Step 1: Pick a point on the terminal ray. For any point

(x, y)

on the terminal ray,

\cot\theta = \frac{x}{y}

. In Quadrant II,

x < 0

and

y > 0

, so use

x = -4

and

y = 3

Step 2: Find its distance from the origin.

r = \sqrt{x^2 + y^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{25} = 5

Step 3: Find cosine and sine, then take reciprocals.

\cos\theta = \frac{x}{r} = -\frac{4}{5} \quad\Rightarrow\quad \sec\theta = -\frac{5}{4}

\sin\theta = \frac{y}{r} = \frac{3}{5} \quad\Rightarrow\quad \csc\theta = \frac{5}{3}

Answer:

\sec\theta = -\frac{5}{4}

and

\csc\theta = \frac{5}{3}

. The signs check out: in Quadrant II, cosine (and so secant) is negative, and sine (and so cosecant) is positive.

Summary

	$\sec\theta$	$\csc\theta$	$\cot\theta$
Reciprocal of	$\cos\theta$	$\sin\theta$	$\tan\theta$
Domain	$\theta \neq \frac{\pi}{2} + k\pi$	$\theta \neq k\pi$	$\theta \neq k\pi$
Range	$(-\infty, -1] \cup [1, \infty)$	$(-\infty, -1] \cup [1, \infty)$	all real numbers
Vertical asymptotes	$\theta = \frac{\pi}{2} + k\pi$	$\theta = k\pi$	$\theta = k\pi$
Zeros	none	none	$\theta = \frac{\pi}{2} + k\pi$
Period	$2\pi$	$2\pi$	$\pi$
Symmetry	even	odd	odd

Here

k

is any integer.

Common mistakes

Mixing up the pairs. $\sec\theta = \frac{1}{\cos\theta}$ and $\csc\theta = \frac{1}{\sin\theta}$ , not the other way around.

Confusing reciprocal with inverse. $\sec\theta$ equals $(\cos\theta)^{-1}$ , but it is not $\cos^{-1}\theta$ , which means arccosine (an angle).

Calling $\cot\frac{\pi}{2}$ undefined. It equals 0. Only the “1 over tangent” form breaks there.

Writing an interval across an asymptote. Secant is increasing on $\left(0, \frac{\pi}{2}\right)$ and on $\left(\frac{\pi}{2}, \pi\right)$ , but not on $(0, \pi)$ .

Forgetting the range gap. An equation like $\sec\theta = \frac{1}{2}$ has no solution, because secant never takes values strictly between $-1$ and 1.

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