Unit 3 · Topic 3.11 · about 30 minutes · 7 parts · 14 scored checks, 4 of them multiple choice

AP Precalculus 3.11: The Secant, Cosecant, and Cotangent Functions

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

The graph of y = csc theta for theta from -1 to 7 radians, drawn as separate U-shaped branches over the dashed graph of y = sin theta. Dashed vertical asymptotes stand at theta = 0, pi and 2pi, where sin theta = 0. The branch between 0 and pi opens upward and touches the sine curve at (pi/2, 1); the branch between pi and 2pi opens downward and touches it at (3pi/2, -1). A shaded band between y = -1 and y = 1 holds no point of the cosecant graph.

Part 1 Think first A beam that runs away

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A lighthouse stands on a rock 2 kilometers out from a long, straight beach, and its beam turns slowly. When the beam points straight at the beach, the light travels 2 kilometers to reach the sand. As the beam swings around toward running parallel to the beach, what happens to the distance the light travels before it hits the sand?

In short. The secant, cosecant and cotangent functions are the reciprocals of cosine, sine and tangent: $\sec θ = \frac{1}{\cos θ}$ , $\csc θ = \frac{1}{\sin θ}$ and $\cot θ = \frac{\cos θ}{\sin θ}$ .

Each one is not defined where its denominator is 0, and its graph has a vertical asymptote there: secant at $\frac{π}{2} + k π$ , cosecant and cotangent at $k π$ .
Secant and cosecant have range $(- \infty, - 1] \cup [1, \infty)$ , because the reciprocal of a number between $- 1$ and 1 lies outside that interval.
Cotangent is 0 where tangent is not defined, and it is decreasing between consecutive asymptotes.

Part 2 Learn it Three short ideas, each checked as you go

2a. Secant and cosecant

Two new trigonometric functions come from flipping the ones you already know:

\sec θ = \frac{1}{\cos θ}, \cos θ \neq 0 \csc θ = \frac{1}{\sin θ}, \sin θ \neq 0

Secant goes with cosine, and cosecant goes with sine. The names cross over, which is exactly why students mix them up. One way to hold it: each pair has exactly one "co" in it.

To evaluate one, find the cosine or sine first, with its sign, then flip it. At $θ = \frac{π}{3}$ , $\cos θ = \frac{1}{2}$ , so $\sec \frac{π}{3} = 2$ . At $θ = \frac{5 π}{4}$ , $\sin θ = - \frac{\sqrt{2}}{2}$ , so $\csc \frac{5 π}{4} = - \frac{2}{\sqrt{2}} = - \sqrt{2}$ . Flipping never changes the sign.

A reciprocal is not an inverse. $\sec θ$ is an output of cosine turned upside down. The inverse cosine, $\cos^{- 1} x$ or $\arccos x$ , takes a value and hands back an angle. A graphing calculator has no secant key, so you type 1 divided by $\cos$ of the angle, never the $\cos^{- 1}$ key.

Quick check · Multiple choice

No calculator. What is the exact value of $\sec \frac{2 π}{3}$ ?

2b. Asymptotes, and a gap in the range

Where cosine is 0 there is nothing to flip, so $\sec θ$ is not defined. That happens at $θ = \frac{π}{2} + k π$ for every integer $k$ . As $\cos θ$ creeps toward 0, its reciprocal blows up: $\frac{1}{0.1} = 10$ and $\frac{1}{0.01} = 100$ . So the graph of $y = \sec θ$ has a vertical asymptote at every $\frac{π}{2} + k π$ . In the same way, the graph of $y = \csc θ$ has a vertical asymptote wherever $\sin θ = 0$ , at $θ = k π$ .

Where cosine is 1 or $- 1$ , flipping changes nothing. So $\sec 0 = 1$ and $\sec π = - 1$ , and at those angles the graphs of cosine and secant touch.

At every other angle where secant is defined, $\cos θ$ is strictly between $- 1$ and 1 and is not 0. Flip a number like that and you get something bigger than 1 or less than $- 1$ , with the same sign. That leaves a gap. The range of both $y = \sec θ$ and $y = \csc θ$ is

(- \infty, - 1] \cup [1, \infty)

The square brackets matter: outputs of exactly 1 and $- 1$ really happen.

The diagram shows this for cosecant. Each branch stays above $y = 1$ or below $y = - 1$ , and it turns around at the point where it touches the sine curve.

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

The graph of $y = \csc θ$ has a vertical asymptote at $θ =$ for every integer $k$ , because there. At $θ = \frac{π}{2}$ it touches the graph of $y = \sin θ$ at height . Its range is .

2c. Cotangent, and the trap in it

The cotangent is the reciprocal of the tangent:

\cot θ = \frac{1}{\tan θ} = \frac{\cos θ}{\sin θ}

Trust the second form. It is defined wherever $\sin θ \neq 0$ , and it tells you what happens at every angle.

Where $\sin θ = 0$ , at $θ = k π$ , the tangent is 0 and the cotangent is not defined. Close to those angles $\cos θ$ is near 1 or $- 1$ while $\sin θ$ is tiny, so the quotient is huge. The graph of $y = \cot θ$ has a vertical asymptote wherever $\tan θ = 0$ .

Now the trap. At $θ = \frac{π}{2}$ the tangent is not defined, and it is tempting to say the cotangent is not defined either. Use the quotient: $\cot \frac{π}{2} = \frac{0}{1} = 0$ . Where tangent has its asymptotes, cotangent has its zeros.

Follow one branch from 0 to $π$ . Just right of 0, $\cot θ$ is a large positive number. It falls through 0 at $\frac{π}{2}$ . Just left of $π$ , $\cos θ$ is near $- 1$ and $\sin θ$ is small and positive, so $\cot θ$ is a large negative number. Cotangent is decreasing between consecutive asymptotes, the opposite of tangent.

Quick check · Decide and justify

No calculator. A classmate says the graph of $y = \cot θ$ has vertical asymptotes at $θ = \frac{π}{2} + k π$ for every integer $k$ . Is the classmate right?

Which reason supports your answer?

Part 3 Watch it done Build cosecant from sine

Watch each step, then do the last one yourself.

Use the graph of $y = \sin θ$ to describe the graph of $y = \csc θ$ for $0 < θ < 2 π$ . Then find $\csc \frac{7 π}{6}$ .

Step 1 · Mark the asymptotes
$\csc θ = \frac{1}{\sin θ}$ is not defined where $\sin θ = 0$ . Inside the interval that is $θ = π$ , and the ends, $θ = 0$ and $θ = 2 π$ , are asymptotes too. Draw dashed vertical lines there.
Step 2 · Mark where the graphs meet
Where $\sin θ$ is 1 or $- 1$ , its reciprocal is the same number. So the graph of cosecant passes through $(\frac{π}{2}, 1)$ and $(\frac{3 π}{2}, - 1)$ , touching the sine curve at each.
Step 3 · Fill in each branch
Between 0 and $π$ sine is positive, so cosecant is too. As sine falls toward 0 near either end, cosecant climbs without bound, so this branch is a U opening upward with its lowest point at $(\frac{π}{2}, 1)$ . One more point helps: $\sin \frac{π}{4} = \frac{\sqrt{2}}{2}$ , so $\csc \frac{π}{4} = \sqrt{2} \approx 1.41$ . Between $π$ and $2 π$ sine is negative, and the branch opens downward with its highest point at $(\frac{3 π}{2}, - 1)$ .
Why this step?
Why can the branch between 0 and $π$ never dip below $y = 1$ ?
On that interval $0 < \sin θ \leq 1$ . The reciprocal of a positive number that is at most 1 is at least 1, so $\csc θ \geq 1$ there, and it equals 1 only at $\frac{π}{2}$ .
Your step · Calculate

No calculator. Finish it. What is $\csc \frac{7 π}{6}$ ?

Your answer

Part 4 Try it Read secant off the cosine graph

The lab has no secant graph, so you build one from cosine, the way the worked example built cosecant from sine. The graph under the circle traces $y = \cos θ$ , and the readouts give $\cos θ$ , $\sin θ$ and $\tan θ$ for the angle you set.

Start at $θ = \frac{π}{3}$ , read $\cos θ$ , and flip it to get $\sec \frac{π}{3}$ . Then drag P slowly toward $\frac{π}{2}$ and watch $\cos θ$ shrink: its reciprocal, the secant, is running off toward an asymptote. Take P once around the circle and list every angle where the $\cos θ$ readout is exactly 0, then every angle where it is 1 or $- 1$ . The first list is where secant has its asymptotes. The second is where the graph of secant touches the graph of cosine.

At $\frac{π}{2}$ the tangent readout says not defined. Use the cosine and sine readouts there to work out $\cot \frac{π}{2}$ , and notice that it is an ordinary number. Then answer the two questions on paper.

Lab

Unit Circle

The radius is fixed at 1, so the horizontal leg of the triangle is $\cos θ$ . Type an angle such as 3pi/2 in the box to land on it exactly.

Open the full Unit Circle lab

Try it · Calculate

No calculator. How many vertical asymptotes does the graph of $y = \sec θ$ have on the interval $0 \leq θ \leq 4 π$ ?

Your answer

Try it · Calculate

Graphing calculator required. What is $\sec 2.5$ , where the angle 2.5 is in radians? Give a decimal to two places.

Your answer

Part 5 Explain it Write it the way the exam asks

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 37.5% of the exam score. This one is shaped like Question 4, Symbolic Manipulations.

Free-response practice · Question 4 style

No calculator. Let $h (θ) = \cot θ \cdot \sec θ$ . (A) Find the exact value of $h (\frac{π}{3})$ . Show the work that leads to your answer. (B) Rewrite $h (θ)$ as an expression in $\sin θ$ and $\cos θ$ only, and simplify it. Then find every value of $θ$ with $0 \leq θ < 2 π$ at which $h$ is not defined. (C) Explain why $h (θ)$ is never strictly between $- 1$ and 1 at any $θ$ where $h$ is defined.

Your answer stays on this page. It is never sent or saved.

Model answer

(A) $\cot \frac{π}{3} = \frac{\cos (π / 3)}{\sin (π / 3)} = \frac{1 / 2}{\sqrt{3} / 2} = \frac{1}{\sqrt{3}}$ and $\sec \frac{π}{3} = \frac{1}{1 / 2} = 2$ , so $h (\frac{π}{3}) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$ . (B) $h (θ) = \frac{\cos θ}{\sin θ} \cdot \frac{1}{\cos θ} = \frac{1}{\sin θ}$ , where $\sin θ \neq 0$ and $\cos θ \neq 0$ . The cotangent is not defined where $\sin θ = 0$ , at $θ = 0$ and $π$ , and the secant is not defined where $\cos θ = 0$ , at $θ = \frac{π}{2}$ and $\frac{3 π}{2}$ . So $h$ is not defined at $0$ , $\frac{π}{2}$ , $π$ and $\frac{3 π}{2}$ . (C) Wherever $h$ is defined, $h (θ) = \frac{1}{\sin θ}$ with $0 < | \sin θ | \leq 1$ . The reciprocal of a number whose absolute value is at most 1 has absolute value at least 1, so $h (θ) \geq 1$ or $h (θ) \leq - 1$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds $\cot \frac{π}{3} = \frac{1}{\sqrt{3}}$ and $\sec \frac{π}{3} = 2$ , and gives $h (\frac{π}{3}) = \frac{2 \sqrt{3}}{3}$ or an equivalent exact value.
Rewrites $h (θ)$ as $\frac{\cos θ}{\sin θ} \cdot \frac{1}{\cos θ}$ and simplifies it to $\frac{1}{\sin θ}$ .
Lists all four of $0$ , $\frac{π}{2}$ , $π$ and $\frac{3 π}{2}$ , keeping the angles where $\cos θ = 0$ even though $\frac{1}{\sin θ}$ is defined there.
Explains that $0 < | \sin θ | \leq 1$ , so its reciprocal has absolute value at least 1.

Part 6 Check yourself Three functions in a table

The three case questions use the same table. Answer them from it.

Each of the functions $p$ , $q$ and $r$ is one of $\sec θ$ , $\csc θ$ and $\cot θ$ , a different one for each. The table gives their values at five angles, in radians.

Values of the functions $p$ , $q$ and $r$
$θ$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$
$p (θ)$	$2$	$\sqrt{2}$	$\frac{2 \sqrt{3}}{3}$	$1$	$\frac{2 \sqrt{3}}{3}$
$q (θ)$	$\sqrt{3}$	$1$	$\frac{\sqrt{3}}{3}$	$0$	$- \frac{\sqrt{3}}{3}$
$r (θ)$	$\frac{2 \sqrt{3}}{3}$	$\sqrt{2}$	$2$	not defined	$- 2$

Case question 1 · Multiple choice

No calculator. Which statement about the functions $p$ , $q$ and $r$ in the table is true?

Case question 2 · Calculate

No calculator. What is $r (\frac{5 π}{3})$ ?

Your answer

Case question 3 · Multiple choice

No calculator. Which number is an output of the function $r$ in the table for some value of $θ$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student describes the graph of $y = \cot θ$ between its vertical asymptotes at $θ = 0$ and $θ = π$ . Tap the first line that goes wrong.

Check yourself · Match

No calculator. Tap each function, then the description of its graph. Here $k$ stands for any integer.

Check yourself · Sort it

No calculator. Each card is an angle in radians. Tap a card, then what is true of $\sec θ$ , $\csc θ$ and $\cot θ$ at that angle.

Only $\sec θ$ is not defined

$\csc θ$ and $\cot θ$ are not defined

All three are defined

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 3.10 Trigonometric Equations and Inequalities · Fill the statement

No calculator. Tap a blank, then tap an angle. The equation is $\sin x = - \frac{1}{2}$ , in radians.

With $0 \leq x < 2 π$ , the solutions are and , smaller first. Every other solution is one of those two plus a whole-number multiple of .

From 3.8 The Tangent Function · Multiple choice

No calculator. The terminal ray of an angle $θ$ in standard position meets the unit circle at $P = (0.8, - 0.6)$ . What is $\tan (θ + π)$ ?

In three lines

$\sec θ = \frac{1}{\cos θ}$ and $\csc θ = \frac{1}{\sin θ}$ : find the cosine or sine with its sign, then flip it.
Secant has vertical asymptotes where $\cos θ = 0$ and cosecant where $\sin θ = 0$ ; both have range $(- \infty, - 1] \cup [1, \infty)$ .
$\cot θ = \frac{\cos θ}{\sin θ}$ has vertical asymptotes at $k π$ , where $\tan θ = 0$ , equals 0 at $\frac{π}{2} + k π$ , and decreases between asymptotes.

Vocabulary

Secant function: The function $f (θ) = \sec θ$ , the reciprocal of the cosine function, defined where $\cos θ \neq 0$ .
Cosecant function: The function $f (θ) = \csc θ$ , the reciprocal of the sine function, defined where $\sin θ \neq 0$ .
Cotangent function: The function $f (θ) = \cot θ$ , the reciprocal of the tangent function where $\tan θ \neq 0$ ; equivalently, $\cot θ = \frac{\cos θ}{\sin θ}$ where $\sin θ \neq 0$ .
Reciprocal: The reciprocal of a nonzero number $x$ is $\frac{1}{x}$ , which has the same sign as $x$ ; a reciprocal is not an inverse function.
Vertical asymptote: A vertical line the graph approaches as its outputs grow without bound; secant and cosecant have them where cosine and sine are zero, respectively, and cotangent where $\tan θ = 0$ .
Range: The set of output values of a function; for secant and cosecant it is $(- \infty, - 1] \cup [1, \infty)$ .
Decreasing: A function is decreasing on an interval when its outputs fall as the input rises; cotangent is decreasing between consecutive asymptotes.

Course alignment, for teachers

AP Precalculus topic 3.11, Unit 3: Trigonometric and Polar Functions.

Skill 2.B: Construct equivalent graphical, numerical, analytical, and verbal representations of functions that are useful in a given mathematical or applied context, with and without technology.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 5 essential knowledge statements of the topic to the part that teaches each one: 5 are checked by a scored question. The map is in the lesson file.

Part 1 Think first A beam that runs away

Part 2 Learn it Three short ideas, each checked as you go

2a. Secant and cosecant

2b. Asymptotes, and a gap in the range

2c. Cotangent, and the trap in it

Part 3 Watch it done Build cosecant from sine

Part 4 Try it Read secant off the cosine graph

Unit Circle

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself Three functions in a table

Part 7 Keep it Bring back earlier ideas

Vocabulary

Labs that use this topic