Read, sketch and remember all four graphs

Explain it like I am five

Picture building a fence: set the posts, hang a panel between each pair, then repeat the section down the line. Sketching tan, cot, sec or csc works the same way.

Posts: write the function as a fraction and find where its bottom is 0; each such input gets a dashed vertical asymptote. Panels: plot a few known points and draw the right shape between posts, rising for tan, falling for cot, U shapes for sec and csc. Repeat: copy by the period, π for tan and cot, 2π for sec and csc.

Example: the posts of sec x = $1 over cos x$ from 0 to 3π. Cosine is 0 at $π over 2$ + nπ, for any integer n. Try n = 0, 1, 2: $π over 2$ , $3 π over 2$ , $5 π over 2$ . The next, $7 π over 2$ , is past 3π, which equals $6 π over 2$ . At the ends cosine is 1 and −1, not 0, so no post there. Three posts.

Nothing here needs memorizing as a list: the bottom of each fraction gives the posts, and the circle gives the points.

In plain words

Picture a route map with landmarks and gaps. Before drawing a route, you mark where it cannot go. Then you add a few known points and connect each allowed piece with its correct shape. The same plan works for all four graphs. Your denominator locates the gaps, your exact values locate the landmarks, and your period tells you how to copy the pattern. The summary below gathers facts you can now explain. Some facts deserve quick recall, but long lists of angles and points can be rebuilt from the circle. Practice rebuilding them instead of trying to memorize every row.

Angles run horizontally in radians; check points and branch boundaries separately.

Reminder

Denominator check. sec( $π over 2$ ) fails because cos( $π over 2$ ) = 0.
Interval endpoints. [−2π, 2π] includes ±2π; (−2π, 2π) would not.
Fractions of π. $π over 2$ + π = $3 π over 2$ .
Exact circle values. At π the point is (−1, 0), so sec π = −1.
Period. tan(x + π) = tan x; csc(x + π) changes sign.
Range. A reciprocal height $1 over 2$ is impossible because −1 < $1 over 2$ < 1.
Symmetry. An odd pair is (x, y) and (−x, −y); secant instead keeps the height.

Why it works. A denominator zero identifies a missing angle input, while nearby signs explain the branch's direction. Exact values anchor the curve so a rough sketch has the right height. The range prevents reciprocal branches from entering the middle strip. Period checks horizontal repetition, and symmetry checks opposite sides of the drawing. These checks use different information, so agreement among them helps catch errors. No single sample point can verify the whole graph, but the combined structure makes a reliable parent-function sketch.

RuleRule: for these four parent graphs, every denominator-zero input is excluded and has unbounded nearby outputs, so its vertical line is a vertical asymptote. Every plotted point must agree with the defining formula and its allowed output range. Period and symmetry describe the repeated and paired points.

The same idea, five ways

Say it

Say build the graph, then check it from several directions.

Write it

Each parent graph must agree with its formula, domain, range, period and symmetry.

In math

cosine-bottom family: tan, sec
sine-bottom family: cot, csc
odd: tan, cot, csc
even: sec
graph words: mark breaks, anchor branches, repeat and check

A route map needs correct gaps, landmarks and repeated sections.

See it

Angles run horizontally in radians; check points and branch boundaries separately.

The same idea, other ways

As a list of forbidden inputs

Read a function's column in the picture. The cosine-bottom pair shares one family, and the sine-bottom pair shares the other. Here n is any integer.

The shared denominator explains the repeated entries.

As a list of heights

The quotient functions reach all real heights. The reciprocal functions stay outside the open strip between −1 and 1.

Match each range to the kind of fraction that defines the function.

As repeating pieces

Quotients copy after one branch of width π. Reciprocal functions need two opposite-sign branches, totaling 2π.

The repeat distances come from the half-turn sign argument.

As opposite-side checks

Secant has equal heights at opposite inputs; the other three have opposite heights.

Even and odd describe the paired points on the graph.

Function	Undefined inputs and vertical asymptotes	Range	Fundamental period	Even or odd
tan x	x = $π over 2$ + nπ	(−∞, ∞)	π	odd
cot x	x = nπ	(−∞, ∞)	π	odd
sec x	x = $π over 2$ + nπ	(−∞, −1] ∪ [1, ∞)	2π	even
csc x	x = nπ	(−∞, −1] ∪ [1, ∞)	2π	odd

.1Know cold, with memory devices

These few facts should come to mind quickly because they start every problem. Keep the denominator check, the quotient and reciprocal pairs, and the two period sizes ready. Use each short memory device with a worked case so the phrase points to a reason instead of replacing it.

Know cold: check the bottom. Bottom 0 means undefined; top 0 over a nonzero bottom means output 0.
Know cold: tan and sec have a cosine bottom; cot and csc have a sine bottom. Memory device: pair by the bottom.
Know cold: quotients cancel two sign changes, so their fundamental period is π; reciprocals carry one sign change and need 2π.
Know cold: even means equal heights; odd means opposite input and opposite output. Sec is even; tan, cot and csc are odd.

Check the bottom

Pair by the bottom

Quotients: π; reciprocals: 2π

Even: equal heights; odd: opposite heights

Each memory device points back to a derivation you have practiced.

The same idea, five ways

Say it

Say the cue, then show why it works. The arrow ⇒ means if the statement on its left holds, then the statement on its right follows.

Write it

The short cues help you start a calculation without memorizing a long angle list.

In math

bottom = 0 ⇒ undefined
top = 0 and bottom ≠ 0 ⇒ output 0
$P_{\tan}$ = $P_{\cot}$ = π
$P_{\sec}$ = $P_{\csc}$ = 2π

A route's few main landmarks are easier to remember than every street address.

See it

cos bottom: tan and sec

sin bottom: cot and csc

The denominator pairing rebuilds the missing inputs.

Worked exampleUse check the bottom

In plain words, apply the same memory cue to two different denominators. Decide whether cot( $π over 2$ ) and sec( $π over 2$ ) have outputs.

Check each function's bottom even when the angle is the same.

cot( $π over 2$ ) = $0 over 1$ = 0.Its sine bottom is nonzero.
sec( $π over 2$ ) would be $1 over 0$ , so it is undefined.Its cosine bottom is zero.

Answer

cot( $π over 2$ ) = 0.
sec( $π over 2$ ) is undefined.

Check The shared circle point (0, 1) supplies both denominators and confirms why the outcomes differ.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: check the bottom means any zero makes the function undefined.

The cue distinguishes the position of the zero.

✓ Instead: has output zero; has no output.

Tips and tricks

Tip: Practice saying the cue and naming its denominator in the same sentence.

.2Understand, then rebuild it when needed

You can rebuild long lists of axis angles, exact values and graph points from a few circle positions. Do not spend memory on a list that takes a few steps to recover. Start with the formula, find the denominator zeros, then use the circle's coordinates for the points between them. Rebuilding also gives you a check when you doubt a remembered answer.

Rebuild forbidden angle lists from cos x = 0 or sin x = 0 and integer multiples of π.
Rebuild exact heights from circle coordinates, then divide or take a reciprocal.
Rebuild asymptote behavior from the denominator's sign near zero and the nonzero numerator.
Rebuild full graphs by copying a period and checking the symmetry; the U-shaped branches come from reciprocal heights.

Angles run horizontally in radians; check points and branch boundaries separately.

The same idea, five ways

Say it

Say start with the formula and rebuild the picture.

Write it

The graph follows from its denominator, a few exact heights and repetition.

In math

tan x = $\frac{\sin x}{\cos x}$
cot x = $\frac{\cos x}{\sin x}$
sec x = $1 over cos x$
csc x = $1 over sin x$

Use a map's main roads to recover the smaller turns.

See it

Angles run horizontally in radians; check points and branch boundaries separately.

Worked exampleRebuilding y = csc x from its downward branch on (−π, 0)

Graph the branch of y = csc x that passes through the point (− $π over 2$ , −1). Then extend the graph using the period and the odd symmetry of cosecant, and state the range. The three-column table gives exact values on this branch: csc(− $5 π over 6$ ) = −2, csc(− $π over 2$ ) = −1 and csc(− $π over 6$ ) = −2. The table lists each input as a count of sixths of π, so −5, −3 and −1 stand for − $5 π over 6$ , − $π over 2$ and − $π over 6$ .

Values of y = csc x on the branch between the asymptotes x = −π and x = 0. The inputs −5, −3 and −1 are counts of sixths of π, standing for −, − and −; the outputs are −2, −1 and −2.

Write the reciprocal formula: y = csc x = $1 over sin x$ .Cosecant is defined as the reciprocal of sine, so sin x is the denominator. Any input that makes sin x equal 0 cannot be used.
Set the denominator equal to 0. sin x = 0 when x = kπ for an integer k. The multiples of π on either side of − $π over 2$ are x = −π and x = 0. Substitute to verify: sin(−π) = 0 and sin(0) = 0. Test just inside the branch: csc(−0.01) = $1 over sin(−0.01)$ ≈ $1 over −0.0099998$ ≈ −100.0. Similarly, sin(−π + 0.01) = −sin(0.01), so csc(−π + 0.01) ≈ −100.0. Draw dashed vertical asymptotes at x = −π and x = 0.Both inputs make the denominator 0, so they are excluded from the domain. Right beside them sin x is a tiny negative number, so its reciprocal is a huge negative number. An excluded input whose nearby outputs are unbounded gives a vertical asymptote. No multiple of π lies strictly between −π and 0, so this branch has no other asymptote.
Convert the table inputs: −5, −3 and −1 sixths of π are − $5 π over 6$ , − $π over 2$ and − $π over 6$ . Confirm each output with the formula. sin(− $5 π over 6$ ) = − $1 over 2$ , so csc(− $5 π over 6$ ) = −2. sin(− $π over 2$ ) = −1, so csc(− $π over 2$ ) = −1. sin(− $π over 6$ ) = − $1 over 2$ , so csc(− $π over 6$ ) = −2. Plot (− $5 π over 6$ , −2), (− $π over 2$ , −1) and (− $π over 6$ , −2).Each plotted point must satisfy y = $1 over sin x$ and stay in the allowed outputs y ≤ −1 or y ≥ 1. All three outputs are at most −1. All three inputs lie strictly between −π and 0, so they belong to the same branch.
Draw a downward-opening U between the asymptotes. Start far below the x-axis just right of x = −π. Rise through (− $5 π over 6$ , −2) to the highest point (− $π over 2$ , −1). Then fall through (− $π over 6$ , −2) and drop toward −∞ as x approaches 0 from the left. The branch is symmetric about x = − $π over 2$ : the inputs − $5 π over 6$ and − $π over 6$ are each $π over 3$ from it and share the output −2.On (−π, 0), sin x is negative. It equals −1 at x = − $π over 2$ and rises toward 0 at both ends. Its reciprocal is therefore at most −1 and reaches −1 only at x = − $π over 2$ . Its size grows without bound as sin x shrinks toward 0.
Repeat by the fundamental period 2π. Adding 2π to each key point gives ( $7 π over 6$ , −2), ( $3 π over 2$ , −1) and ( $11 π over 6$ , −2), on a copy between x = π and x = 2π. Subtracting 2π gives a copy between x = −3π and x = −2π with top (− $5 π over 2$ , −1). Next, check the odd symmetry: csc(−x) = $1 over sin(−x)$ = $1 over −sin x$ = −csc x, so each point (a, b) pairs with (−a, −b). Reflecting the key points through the origin gives ( $5 π over 6$ , 2), ( $π over 2$ , 1) and ( $π over 6$ , 2). These form an upward U on (0, π) with lowest point ( $π over 2$ , 1), which also repeats every 2π. Range: (−∞, −1] ∪ [1, ∞).Sine has period 2π, so csc(x + 2π) = $1 over sin(x + 2 π)$ = $1 over sin x$ = csc x, and shifted points stay on the graph. Sine is odd, so its reciprocal is odd and the graph is symmetric about the origin. Because |sin x| ≤ 1, every output has |csc x| ≥ 1, so no point lies strictly between y = −1 and y = 1.

Answer

The vertical asymptotes x = −π and x = 0 bound the branch. It is a downward-opening U with highest point (−, −1). It passes through (−, −2) and (−, −2) and falls toward −∞ beside both asymptotes. By odd symmetry, the next branch is an upward U on (0, π) with lowest point (, 1), passing through (, 2) and (, 2). Repeating every 2π gives the whole graph: asymptotes x = kπ (k any integer), tops (− + 2kπ, −1), bottoms ( + 2kπ, 1), and range (−∞, −1] ∪ [1, ∞).

Check Test an input that is not in the table. x = − lies between − and −, so the sketch puts its output between −1 and −2. The formula gives sin(−) = −, so csc(−) = − = − ≈ −1.41, which is between −2 and −1. Period: sin() = −1, so csc() = −1, matching the shifted top (, −1). Symmetry: sin() = , so csc() = 2, the opposite of csc(−) = −2. Asymptote: csc(−0.01) ≈ −100.0, so the branch does plunge beside x = 0.

Work to write

y = csc x = $1 over sin x$
sin x = 0 ⇒ x = kπ; sin(−π) = 0 and sin(0) = 0
vertical asymptotes x = −π and x = 0 (csc(−0.01) ≈ −100)
points (− $5 π over 6$ , −2), (− $π over 2$ , −1), (− $π over 6$ , −2), all with y ≤ −1
downward U on (−π, 0), highest point (− $π over 2$ , −1), y → −∞ at both asymptotes
period 2π: copy on (π, 2π) through ( $7 π over 6$ , −2), ( $3 π over 2$ , −1), ( $11 π over 6$ , −2)
odd: csc(−x) = −csc x ⇒ upward U on (0, π), lowest point ( $π over 2$ , 1)
range (−∞, −1] ∪ [1, ∞)

The vertical asymptotes x = −π and x = 0 bound the branch. It is a downward-opening U with highest point (− $π over 2$ , −1). It passes through (− $5 π over 6$ , −2) and (− $π over 6$ , −2) and falls toward −∞ beside both asymptotes. By odd symmetry, the next branch is an upward U on (0, π) with lowest point ( $π over 2$ , 1), passing through ( $π over 6$ , 2) and ( $5 π over 6$ , 2). Repeating every 2π gives the whole graph: asymptotes x = kπ (k any integer), tops (− $π over 2$ + 2kπ, −1), bottoms ( $π over 2$ + 2kπ, 1), and range (−∞, −1] ∪ [1, ∞).

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: memorizing sec π = 1 from a picture is enough.

A picture can be misread, and cosine at π is −1.

✓ Instead: Rebuild sec π = 1 ÷ (−1) = −1.

Tips and tricks

Tip: Do not memorize every point. Keep the circle and the four formulas ready to rebuild them.

.3Put on the cheat sheet: printable summary

This compact card is the information to look up while studying or to put on an allowed cheat sheet. The longer derivations above tell you why each entry is true. On a closed-book exam you can recreate the card from the denominator pairs, the circle and the period argument. Here n always means any integer, including negative integers and zero.

tan: $\frac{\sin x}{\cos x}$ . Exclude $π over 2$ + nπ. Range (−∞, ∞). Fundamental period π. Odd; each branch rises.
cot: $\frac{\cos x}{\sin x}$ . Exclude nπ. Range (−∞, ∞). Fundamental period π. Odd; each branch falls.
sec: $1 over cos x$ . Exclude $π over 2$ + nπ. Range (−∞, −1] ∪ [1, ∞). Fundamental period 2π. Even; reciprocal of cosine.
csc: $1 over sin x$ . Exclude nπ. Range (−∞, −1] ∪ [1, ∞). Fundamental period 2π. Odd; reciprocal of sine.

tan: cos bottom; P = π; odd

cot: sin bottom; P = π; odd

sec: cos bottom; P = 2π; even

csc: sin bottom; P = 2π; odd

Read the accompanying four facts for the complete domain and range information.

The same idea, five ways

Say it

Say use the card for a quick structure check.

Write it

The summary gathers the four parent graphs in one place.

In math

cosine zeros: $π over 2$ + nπ
sine zeros: nπ
quotient range: (−∞, ∞)
reciprocal range: (−∞, −1] ∪ [1, ∞)

A small map key helps you read a larger map.

See it

The key organizes the four functions into two families.

Worked exampleCheck a reciprocal sketch using the card

A sketch labeled csc x has a branch passing through (π, 0). Is that possible? In plain words, check one claimed point against both domain and range.

Angles run horizontally in radians; check points and branch boundaries separately.

Input π is excluded because sin π = 0.Setting the denominator equal to zero locates a forbidden input, and substitution verifies it.
Output 0 is outside the range of cosecant.Every valid cosecant output is at least 1 or at most −1.
Replace the claimed point with the asymptote x = π and keep each neighboring branch away from the x-axis.The graph approaches that input with unbounded outputs; it never has an output there.

Answer

The point (π, 0) is impossible.
Draw the vertical asymptote x = π instead.

Check Two independent checks fail: the proposed input is forbidden and the proposed output is in the missing range band.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: a U-shaped secant branch can dip to y = .

The summary's reciprocal range rejects that height.

✓ Instead: The positive branch has its lowest height at 1.

Tips and tricks

Tip: A summary card is a check, not a substitute for the denominator and reciprocal explanations.

Strategy: step by step

1. Write the quotient or reciprocal formula.
2. Set its denominator equal to 0 to locate the asymptotes; substitute a candidate back to verify the zero.
3. Plot the known key points within each allowed branch.
4. Draw the rising tangent, falling cotangent, or reciprocal U-shaped branches.
5. Repeat by the fundamental period and check the even or odd symmetry and the range.

Strategy

Strategy: sketch a parent graph and audit the result

Is this a secant or cosecant graph?

YesUse the corresponding cosine or sine guide; keep reciprocal branches outside the middle band.

NoUse three ratio anchors and the rising tangent or falling cotangent shape.

↓

Would a drawn line cross an asymptote?

YesStop it and start a separate branch on the other side.

NoCheck period and symmetry.

1. Write its defining formula and find the forbidden denominator-zero inputs.
2. Substitute into the denominator to confirm every marked asymptote.
3. Place the known key points and use the correct branch shape.
4. Repeat by its fundamental period.
5. Check its range and its even or odd point pairs.

Worked exampleSecant: the upward branch through (2π, 1) and its neighbor

Graph the branch of y = sec x that passes through the point (2π, 1), then sketch the neighboring downward branch through (3π, −1). Extend the graph using the period and the even symmetry of secant, and state the range. The three-column table gives exact values on the first branch: sec( $7 π over 4$ ) = $square root of 2$ , sec(2π) = 1 and sec( $9 π over 4$ ) = $square root of 2$ . The table lists each input as a multiple of π, so 1.75, 2 and 2.25 stand for $7 π over 4$ , 2π and $9 π over 4$ , and it shows $square root of 2$ rounded to 1.414.

Three-column table of y = sec x on the branch through (2π, 1). The inputs are multiples of π, so 1.75, 2 and 2.25 stand for , 2π and . The outputs are 1.414 ( rounded to three decimals), 1 and 1.414. The inputs 1.75 and 2.25 are the paired points either side of 2π.

Write secant as a reciprocal formula: y = sec x = $1 over cos x$ .Secant is the reciprocal of cosine, so each output is 1 divided by cos x. The formula has a value only where its denominator cos x is not 0.
Set the denominator equal to 0: cos x = 0 gives x = $π over 2$ + kπ for any integer k. The two zeros on either side of 2π are x = $3 π over 2$ and x = $5 π over 2$ . Substitute them back: cos( $3 π over 2$ ) = 0, and cos( $5 π over 2$ ) = cos( $π over 2$ + 2π) = cos( $π over 2$ ) = 0.A denominator-zero input is excluded. Just beside it cos x is close to 0, so $1 over cos x$ grows without bound, which makes each such vertical line a vertical asymptote. Consecutive zeros are π apart, so no other asymptote lies between $3 π over 2$ and $5 π over 2$ , and these two lines bound the branch through (2π, 1).
Find the allowed outputs of this branch. For $3 π over 2$ < x < $5 π over 2$ , the angle x − 2π is between − $π over 2$ and $π over 2$ , so 0 < cos x ≤ 1. Therefore sec x = $1 over cos x$ ≥ 1, and sec x = 1 only where cos x = 1, which is at x = 2π.The reciprocal of a positive number no greater than 1 is at least 1. So every point of this branch must lie on or above the line y = 1, and the branch opens upward.
Verify and plot the table points. cos(2π) = 1 gives sec(2π) = 1. cos( $7 π over 4$ ) = $square root of 2 over 2$ gives sec( $7 π over 4$ ) = $2 over square root of 2$ = $square root of 2$ ≈ 1.414. cos( $9 π over 4$ ) = cos( $π over 4$ + 2π) = $square root of 2 over 2$ gives sec( $9 π over 4$ ) = $square root of 2$ . Plot (2π, 1), ( $7 π over 4$ , $square root of 2$ ) and ( $9 π over 4$ , $square root of 2$ ), at x ≈ 6.28, 5.50 and 7.07.Each plotted point must agree with sec x = $1 over cos x$ and with the allowed outputs y ≥ 1. The two $square root of 2$ points sit $π over 4$ either side of 2π at the same height, so the branch is symmetric about the line x = 2π.
Draw the upward U. Start high beside x = $3 π over 2$ , fall through ( $7 π over 4$ , $square root of 2$ ) to the lowest point (2π, 1), rise through ( $9 π over 4$ , $square root of 2$ ), and climb beside x = $5 π over 2$ without ever touching either asymptote.From x = $3 π over 2$ to x = 2π, cos x increases from 0 to 1, so its reciprocal falls from unbounded values down to 1. From x = 2π to x = $5 π over 2$ , cos x decreases from 1 toward 0, so sec x increases without bound. Near both asymptotes cos x approaches 0 through positive values, so the outputs head to +∞.
Sketch the neighboring branch. The next zero of cos x is x = $7 π over 2$ , since cos( $7 π over 2$ ) = cos( $3 π over 2$ + 2π) = 0. For $5 π over 2$ < x < $7 π over 2$ , −1 ≤ cos x < 0. cos 3π = cos(π + 2π) = −1 gives sec 3π = −1. cos( $11 π over 4$ ) = cos( $13 π over 4$ ) = − $square root of 2 over 2$ gives sec( $11 π over 4$ ) = sec( $13 π over 4$ ) = − $square root of 2$ . Draw a downward U between x = $5 π over 2$ and x = $7 π over 2$ through ( $11 π over 4$ , − $square root of 2$ ), the highest point (3π, −1), and ( $13 π over 4$ , − $square root of 2$ ).The reciprocal of a negative number between −1 and 0 is at most −1, so this branch lies on or below y = −1. Near its asymptotes cos x approaches 0 through negative values, so the outputs fall without bound.
Repeat by the fundamental period 2π. The window from x = $3 π over 2$ to x = $7 π over 2$ holds one upward and one downward branch, and copies of it shifted by 2π, 4π, … in both directions give the whole graph. For example, shifting the first branch left by 2π gives the upward branch between x = − $π over 2$ and x = $π over 2$ through (− $π over 4$ , $square root of 2$ ), (0, 1) and ( $π over 4$ , $square root of 2$ ). For integers k, the asymptotes are x = $π over 2$ + kπ, the lowest points of the upward branches are (2kπ, 1), and the highest points of the downward branches are ((2k + 1)π, −1).cos(x + 2π) = cos x, so sec(x + 2π) = sec x and every point (x, y) repeats at (x + 2π, y). The period is 2π and not π because sec(x + π) = −sec x, so a shift of π carries an upward branch onto a downward one.
Check the even symmetry. sec(−x) = $1 over cos(−x)$ = $1 over cos x$ = sec x, so the graph is symmetric about the y-axis. The mirror image of ( $7 π over 4$ , $square root of 2$ ) is (− $7 π over 4$ , $square root of 2$ ), and cos(− $7 π over 4$ ) = $square root of 2 over 2$ gives sec(− $7 π over 4$ ) = $square root of 2$ . The period gives the same value: − $7 π over 4$ = $9 π over 4$ − 4π, and sec( $9 π over 4$ ) = $square root of 2$ .Even symmetry pairs each point (x, y) with (−x, y). The paired point agrees with both the formula and the period, so the extended graph is consistent. The copy through (0, 1) is its own mirror image.
State the range. Wherever sec x is defined, either 0 < cos x ≤ 1, giving sec x ≥ 1 (the upward branches), or −1 ≤ cos x < 0, giving sec x ≤ −1 (the downward branches). The range is (−∞, −1] ∪ [1, ∞).cos x always lies between −1 and 1, so no output of sec x lies strictly between −1 and 1. The values 1 and −1 are reached at x = 2π and x = 3π, and the branches climb or fall without bound near the asymptotes.

Answer

The branch through (2π, 1) is an upward U between the vertical asymptotes x = and x = . Its lowest point is (2π, 1), and it passes through (, ) and (, ). Beside it is a downward U between x = and x = with highest point (3π, −1). The graph repeats every 2π, has vertical asymptotes x = + kπ for every integer k, and is symmetric about the y-axis because sec(−x) = sec x. The range is (−∞, −1] ∪ [1, ∞).

Check Test an input that is not in the table. x = lies between and 2π, and cos() = , so sec() = = ≈ 1.155. This lies between 1 and ≈ 1.414, as the falling left half of the U requires. Just right of the asymptote, x = + 0.01 gives cos x = sin 0.01 ≈ 0.0100 and sec x ≈ 100, a very large positive output. The pair (, ) and (−, ) agrees with both the even symmetry and the period 2π. Every plotted output (1, , −1 and −) satisfies y ≥ 1 or y ≤ −1.

Work to write

sec x = $1 over cos x$
cos x = 0 gives x = $π over 2$ + kπ; check: cos( $3 π over 2$ ) = 0 and cos( $5 π over 2$ ) = 0
Vertical asymptotes of the branch: x = $3 π over 2$ and x = $5 π over 2$
sec(2π) = 1, sec( $7 π over 4$ ) = $2 over square root of 2$ = $square root of 2$ , sec( $9 π over 4$ ) = $square root of 2$
Upward U with lowest point (2π, 1); sec x ≥ 1 on this branch
Downward U between x = $5 π over 2$ and x = $7 π over 2$ with highest point (3π, −1)
Period 2π: sec(x + 2π) = sec x; asymptotes x = $π over 2$ + kπ
Even: sec(−x) = sec x, so sec(− $7 π over 4$ ) = $square root of 2$
Range: (−∞, −1] ∪ [1, ∞)

The branch through (2π, 1) is an upward U between the vertical asymptotes x = $3 π over 2$ and x = $5 π over 2$ . Its lowest point is (2π, 1), and it passes through ( $7 π over 4$ , $square root of 2$ ) and ( $9 π over 4$ , $square root of 2$ ). Beside it is a downward U between x = $5 π over 2$ and x = $7 π over 2$ with highest point (3π, −1). The graph repeats every 2π, has vertical asymptotes x = $π over 2$ + kπ for every integer k, and is symmetric about the y-axis because sec(−x) = sec x. The range is (−∞, −1] ∪ [1, ∞).

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: tangent's zeros x = nπ are its asymptotes.

At those inputs, cosine is ±1 and the quotient is defined with output zero.

✓ Instead: Its asymptotes are the shifted family x = + nπ.

✗ Not this: Counterexample: join a negative reciprocal branch to a positive one through y = 0.

You would cross a forbidden input and enter the missing range band.

✓ Instead: Keep the branches separate, each facing away from the x-axis.

Tips and tricks

Tip: Before sketching, mark forbidden inputs. Before accepting the sketch, check its heights and repeat distance.
Tip: In the summary, n is any integer; +nπ includes every angle in the notation ±nπ because negative integers supply the minus cases.

Trap. Taking the summary as four arbitrary rows. Rebuild each excluded-input family from its denominator and each period from what a half-turn does to the signs.

Keep in mind

On a one-page handwritten cheat sheet, the summary table (asymptotes, range, period, even or odd) and a thumbnail of each graph cover this section, and the rest rebuilds from the four fractions.
Test the endpoints of a closed interval too: on [0, 3π], cot x has posts at both ends, since sine is 0 at 0 and at 3π.
Tangent's zeros, x = nπ, are not its posts; its posts sit halfway between, at $π over 2$ + nπ.
Check a finished sketch three ways: sec and csc never enter −1 < y < 1, the pattern repeats at the right period, and only sec is symmetric about the y-axis.

Memory hookPosts, panels, repeat: bottom zeros are posts, key points and shape make panels, the period repeats them. Tan climbs, cot slides, sec and csc cup.

Flash cards: say the answer out loud, then flip

What is a branch?

One unbroken piece of the graph between two neighboring asymptotes.

Which two functions share the asymptotes x = nπ?

cot x and csc x: both divide by sin x.

Which of the four graphs is symmetric about the y-axis?

sec x, the only even one.

List the asymptotes of csc x on [π, 3π].

x = π, 2π and 3π. The endpoints count, since sine is 0 there.

List the asymptotes of tan x on [4π, 6π].

x = and x = .

True or false: tangent's asymptotes sit at its zeros.

False: its zeros are at nπ, and its asymptotes are at + nπ.