Four formulas and the inputs they must skip

Explain it like I am five

Picture sharing 1 pizza: 2 friends each get $1 over 2$ , but with 0 friends there is no answer, since no share times 0 makes 1. A fraction with 0 on the bottom, the denominator, is undefined: it has no value.

Each function is a fraction built from the circle point for the angle x, (cos x, sin x): tan x = $\frac{\sin x}{\cos x}$ , cot x = $\frac{\cos x}{\sin x}$ , sec x = $1 over cos x$ and csc x = $1 over sin x$ .

Example: x = $5 π over 2$ , a full lap (2π) plus a quarter lap, ends straight up at (0, 1), so cosine is 0 and sine is 1. Then cot $5 π over 2$ = $0 over 1$ = 0 and csc $5 π over 2$ = 1 are fine, but tan $5 π over 2$ and sec $5 π over 2$ both equal $1 over 0$ : undefined.

Cosine is 0 straight up and down, at x = $π over 2$ + nπ for any integer n (a whole number: positive, negative or zero), so tan and sec skip those inputs. Sine is 0 far right and far left, at x = nπ, so cot and csc skip those.

In plain words

Think of four number machines that all begin with the same angle. Each machine first reads the angle's position on a circle. Then it divides using those coordinates. A denominator is the number on the bottom of a fraction. If that number is zero, the machine cannot give an output. The Domain is the set of inputs the machine accepts. A Quadrantal angle ends on an axis, so one of its circle coordinates is zero. You will find which of the four machines uses that zero as a denominator. This gives a reason for every missing input instead of four unrelated lists to memorize. Tangent, cotangent, secant and cosecant are the full names of tan, cot, sec and csc.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Reminder

Unit-circle coordinates. Cosine first, sine second: at $π over 2$ the point (0, 1) gives cos = 0 and sin = 1.
Zero numerator versus denominator. $0 over 5$ = 0, but $5 over 0$ is undefined.
Integer multiples. An integer can be negative or zero: nπ with n = −2 is −2π.
Substitution. Check an excluded input in the bottom: sin π = 0.

Why it works. A quotient is a division. The Quotient identities are equalities writing tangent and cotangent as divisions of sine and cosine. The Reciprocal identities write secant and cosecant as one divided by cosine or sine. An identity is an equality true for all inputs where its two sides exist. Cosine is zero at vertical-axis angles, + nπ. Sine is zero at horizontal-axis angles, nπ. In each family the other coordinate is 1 or −1, so the numerator stays nonzero. Thus each formula fails precisely at its stated denominator zeros.

RuleRule: tan x = ; cot x = ; sec x = ; csc x = .
For tan and sec, require x ≠ + nπ; for cot and csc, require x ≠ nπ, with n any integer.

The same idea, five ways

Say it

Say the domain is every accepted input. The label $D_{\tan}$ names tangent's domain, and $D_{\cot}$ , $D_{\sec}$ and $D_{\csc}$ name the other domains. A subscript is a small label identifying which function the letter belongs to.

Write it

Each function accepts every real angle except the angles that make its denominator zero.

In math

$D_{\tan}$ = $D_{\sec}$ = {x | x ≠ $π over 2$ + nπ, for every integer n}
$D_{\cot}$ = $D_{\csc}$ = {x | x ≠ nπ, for every integer n}
graph words: skip the vertical lines at those input positions

A machine has a list of inputs it cannot accept.

See it

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

The same idea, other ways

As a machine

The machine accepts an angle only if the division stage has a nonzero bottom.

A zero cosine stops the secant machine.

As a circle

Vertical-axis points have cosine 0. Horizontal-axis points have sine 0. Read the denominator to decide which pair stops.

The zero sine at π stops cotangent and cosecant.

With multiplication

For 1 ÷ 0 to have an answer, answer × 0 would need to equal 1. It never can.

No number of zero-sized groups can give a total of one.

As a memory cue

Tip: Check the bottom. Cosine bottom means the vertical axis is forbidden; sine bottom means the horizontal axis is forbidden.

tan, sec: cosine bottom

cot, csc: sine bottom

Bottom 0 means no output

Two denominator families replace four separate lists.

.1Tangent: sine divided by cosine

Tangent compares the vertical coordinate with the horizontal coordinate of a point on the circle. Think of rise divided by run on a ramp. A zero rise can give zero steepness, but a zero run cannot sit on the bottom of a fraction. That is why tangent skips the axis angles where cosine is zero.

Rule: tan x = $\frac{\sin x}{\cos x}$ , with cos x ≠ 0.
Domain: all real x except x = $π over 2$ + nπ, with n any integer.
The same excluded angles in degrees are 90° + 180°n.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

The same idea, five ways

Say it

Say tangent is sine divided by cosine.

Write it

A tangent input is allowed when its cosine is not zero.

In math

tan x = $\frac{\sin x}{\cos x}$
cos x ≠ 0
x ≠ $π over 2$ + nπ
{x | x ≠ $π over 2$ + nπ, for every integer n}

A ramp's rise can be zero; its run cannot be zero in a rise-over-run fraction.

See it

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Worked exampleTangent accepts a zero numerator

In plain words, enter angle zero and decide which output the formula gives. Find tan 0.

At input 0 radians, tangent has the allowed output 0.

At 0 the circle point is (1, 0), so sin 0 = 0 and cos 0 = 1.Cosine is the horizontal coordinate and sine the vertical coordinate.
tan 0 = $0 over 1$ = 0.The denominator is nonzero, so this division has an answer.

Answer

tan 0 = 0.

Check Multiplication checks the division: 0 × 1 = 0.

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

✗ Not this: Counterexample: every quadrantal angle makes tangent undefined.

At x = 0 the denominator is 1, so nothing prevents division.

✓ Instead: Only quadrantal angles on the vertical axis make cosine zero and exclude tangent.

Tips and tricks

Tip: Tangent and secant share the cosine bottom, so they share forbidden inputs.

.2Cotangent: cosine divided by sine

Cotangent reverses tangent's two coordinate roles. It is run divided by rise. The vertical coordinate, sine, now sits on the bottom, so the forbidden inputs move to the horizontal axis. Use this quotient directly whenever tangent itself is undefined. The formula 1 divided by tangent can only be used when tangent exists and is nonzero.

Rule: cot x = $\frac{\cos x}{\sin x}$ , with sin x ≠ 0.
Domain: all real x except x = nπ, with n any integer.
The excluded angles in degrees are 180°n.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

The same idea, five ways

Say it

Say cotangent is cosine divided by sine.

Write it

A cotangent input is allowed when its sine is not zero.

In math

cot x = $\frac{\cos x}{\sin x}$
sin x ≠ 0
x ≠ nπ
{x | x ≠ nπ, for every integer n}

Run divided by rise needs a nonzero rise.

See it

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Worked exampleCotangent rejects angle zero

In plain words, test whether angle zero has a cotangent output. Find cot 0.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

At 0, cos 0 = 1 and sin 0 = 0, so cot 0 would be $1 over 0$ .The quotient places sine on the bottom.
There is no output at x = 0.No number times 0 gives 1. This makes zero a forbidden input.

Answer

cot 0 is undefined.

Check The horizontal-axis circle point (1, 0) has a zero sine, exactly the bottom that cotangent requires.

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

✗ Not this: Counterexample: cot() is undefined because tan() is undefined.

The quotient for cotangent has sine, not cosine, in its denominator.

✓ Instead: cot() = = 0.

Tips and tricks

Tip: Use cos divided by sin to evaluate cotangent at an axis angle.

.3Secant: reciprocal of cosine

Secant asks how many copies of the cosine number make one. If cosine is half, two copies make one, so secant is 2. If cosine is zero, no copies can make one. This makes secant fail at exactly the same angle inputs as tangent because both put cosine on the bottom.

Rule: sec x = $1 over cos x$ , with cos x ≠ 0.
Domain: all real x except x = $π over 2$ + nπ.
The excluded angles in degrees are 90° + 180°n.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

The same idea, five ways

Say it

Say secant is one divided by cosine.

Write it

Secant exists exactly where cosine is nonzero.

In math

sec x = $1 over cos x$
x ≠ $π over 2$ + nπ

How many pieces the size of cosine fit into one whole?

See it

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Worked exampleA negative cosine gives a negative secant

In plain words, take the reciprocal of cosine at a half-turn. Find sec π.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

At π the unit-circle point is (−1, 0), so cos π = −1.Cosine is the first coordinate.
sec π = $1 over −1$ = −1.A negative denominator gives a negative reciprocal.

Answer

sec π = −1.

Check cos π × sec π = (−1) × (−1) = 1.

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

✗ Not this: Counterexample: sec x means cos x with a minus sign.

A reciprocal changes size; a minus sign changes sign. At x = 0 cosine and secant are both 1.

✓ Instead: Compute 1 divided by the nonzero cosine.

Tips and tricks

Tip: The letters in sec and cos differ; remember the pair by saying secant belongs to cosine.

.4Cosecant: reciprocal of sine

Cosecant asks for one divided by sine. If sine is half, its reciprocal is 2. If sine is negative, its reciprocal stays negative. If sine is zero, there is no reciprocal. The forbidden inputs are therefore the same as cotangent's: the angles on the horizontal axis, where the circle point has height zero.

Rule: csc x = $1 over sin x$ , with sin x ≠ 0.
Domain: all real x except x = nπ.
The excluded angles in degrees are 180°n.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

The same idea, five ways

Say it

Say cosecant is one divided by sine.

Write it

Cosecant exists exactly where sine is nonzero.

In math

csc x = $1 over sin x$
x ≠ nπ

Use the height as the size of each piece when counting pieces in one whole.

See it

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Worked exampleCosecant exists on the upper vertical axis

In plain words, enter the quarter-turn and take the reciprocal of its sine. Find csc( $π over 2$ ).

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

At $π over 2$ the point is (0, 1), so sin( $π over 2$ ) = 1.Sine reads the vertical coordinate.
csc( $π over 2$ ) = $1 over 1$ = 1.The bottom is nonzero, so the input is accepted.

Answer

csc() = 1.

Check Its sine partner times it gives 1 × 1 = 1.

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

✗ Not this: Counterexample: csc 0 = 0 because sin 0 = 0.

A zero output of sine becomes a zero denominator for cosecant.

✓ Instead: csc 0 is undefined.

Tips and tricks

Tip: Cosecant belongs to sine; cotangent shares its sine bottom.

Strategy: step by step

1. Write the formula as a fraction because the denominator identifies the restriction.
2. Set the denominator equal to 0 to find the inputs that would make the division impossible.
3. Read those angles from the unit circle and substitute them back into the denominator to verify it is 0.
4. Exclude those inputs. Keep a zero numerator when the denominator is nonzero.

Strategy

Strategy: find the denominator's forbidden angles

Is the denominator cosine?

YesExclude + nπ; for n = 0, cos() = 0 verifies it.

NoThe denominator is sine; exclude nπ, verified by sin 0 = 0.

↓

Is zero only in the numerator?

YesKeep the input; the output is zero when the bottom is nonzero.

NoIf the bottom is zero, the input has no output.

1. Identify the denominator in the quotient or reciprocal formula.
2. Set that denominator equal to 0 to locate failures.
3. Find the full integer family and check a member in the denominator.
4. Exclude that family while keeping valid zero outputs.
5. To list forbidden inputs inside an interval, put the angle family between the two given bounds. Divide all parts by positive π, then subtract $1 over 2$ from all parts if the family is $π over 2$ + nπ. Keep every integer between the resulting bounds on n. This finds exactly the family members inside the requested interval, even if the interval is far from zero. For [−π, 3π] and the cosine-zero family, this gives − $3 over 2$ ≤ n ≤ $5 over 2$ , so n = −1, 0, 1, 2. Compare the coefficients of π to check neighbors: − $3 π over 2$ = −1.5π < −π.

Worked exampleWhich machines accept a quarter-turn?

In plain words, use one angle to decide which machines have a number to return. Find tan, cot, sec and csc at x = $π over 2$ .

The zero horizontal coordinate excludes tan and sec; the height 1 allows cot and csc.

The point at $π over 2$ is (0, 1), so cos( $π over 2$ ) = 0 and sin( $π over 2$ ) = 1.The first coordinate is cosine and the second is sine.
tan( $π over 2$ ) = $1 over 0$ and sec( $π over 2$ ) = $1 over 0$ , so both are undefined.Their common denominator, cosine, is 0. This verifies why the angle is excluded.
cot( $π over 2$ ) = $0 over 1$ = 0 and csc( $π over 2$ ) = $1 over 1$ = 1.Their denominator, sine, is 1, so both divisions are permitted.

Answer

tan( $π over 2$ ): undefined.
cot( $π over 2$ ): 0.
sec( $π over 2$ ): undefined.
csc( $π over 2$ ): 1.

Check At a vertical-axis angle, the two cosine-bottom functions fail and the two sine-bottom functions remain defined, exactly as the rule predicts.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: zero on top

In plain words, decide whether a zero sine causes trouble. Find tan 0.

At input 0 radians, tangent has the allowed output 0.

sin 0 = 0 and cos 0 = 1.The point is (1, 0).
tan 0 = $0 over 1$ = 0.The bottom is 1, so the division is allowed.

Answer

Check 0 × 1 = 0 checks the division.

Rung 2Rung 2: zero on the bottom

In plain words, decide whether a half-turn has a cosecant output. Find csc π.

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

sin π = 0, so csc π would be $1 over 0$ .The point at π is (−1, 0).
Exclude x = π.This is a verified sine zero, so the fraction has no output.

Answer

csc π is undefined.

Check π = nπ with n = 1, so the input belongs to the sine-zero family.

Rung 3Rung 3: a negative axis angle

In plain words, turn clockwise three quarters of a full turn and evaluate both fractions. Find sec(− $3 π over 2$ ) and cot(− $3 π over 2$ ).

The clockwise turn ends on the upper vertical axis.

− $3 π over 2$ ends at (0, 1), so cosine is 0 and sine is 1.Adding a full turn 2π reaches $π over 2$ , the same circle point.
sec = $1 over 0$ is undefined; cot = $0 over 1$ = 0.Use the actual denominators, even for a negative input.

Answer

sec: undefined.
cot: 0.

Check − = + (−2)π, and substituting this angle into cosine indeed gives 0.

Rung 4Rung 4: a whole interval of exclusions

In plain words, find every cosine-zero angle in this closed interval. List the forbidden secant inputs in [−π, 3π].

The horizontal inputs are radians; the missing inputs separate the graph into pieces.

Use x = $π over 2$ + nπ because secant divides by cosine.Setting cos x = 0 locates the inputs to discard.
Divide −π ≤ $π over 2$ + nπ ≤ 3π by positive π to get −1 ≤ $1 over 2$ + n ≤ 3. Subtract $1 over 2$ from all three parts: − $3 over 2$ ≤ n ≤ $5 over 2$ .This finds exactly which integer choices make the excluded angle lie in the requested interval. Positive division preserves order and equal subtraction preserves the bounds.
Substitute n = −1, 0, 1, 2 to obtain − $π over 2$ , $π over 2$ , $3 π over 2$ , $5 π over 2$ . Each angle's cosine is 0.These are the vertical-axis angles inside the stated boundaries.
The next angles outward are − $3 π over 2$ < −π and $7 π over 2$ > 3π, so leave them out.Checking the neighboring candidates proves the list is complete.

Answer

x = − $π over 2$ .
x = $π over 2$ .
x = $3 π over 2$ .
x = $5 π over 2$ .

Check The endpoints −π and 3π have cosine −1, so neither endpoint is forbidden.

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

✗ Not this: Counterexample: tan π is undefined because π is quadrantal.

At π the zero is sine, on top. Cosine is −1, on the bottom.

✓ Instead: tan π = = 0.

✗ Not this: Counterexample: n must be positive in x = nπ.

The notes use an integer n; negative integers and zero are included.

✓ Instead: n = −1 gives −π, n = 0 gives 0, and both are sine zeros.

Tips and tricks

Tip: Write the fraction before marking a forbidden input.
Tip: Keep the memory cue check the bottom beside your domain rule.

Trap. A zero on the top is not a break. Check the bottom first: = 0, while is undefined.

Keep in mind

Write the fraction first, then ask where its bottom is 0: sec x = $1 over cos x$ , and cos x = 0 at $π over 2$ , − $π over 2$ , $5 π over 2$ and so on.
The n in $π over 2$ + nπ may be negative: n = −3 gives $π over 2$ − 3π = − $5 π over 2$ .
A 0 on top is allowed: cot x = 0 wherever cos x = 0, because the bottom, sin x, is then 1 or −1.
In this section x names the angle, the graph's input, not the circle point's x-coordinate: at x = $π over 2$ the point is (0, 1), so cos x = 0.

Memory hookCo names divide by sine, plain names by cosine: cotangent and cosecant skip x = nπ; tangent and secant skip x = + nπ.

Flash cards: say the answer out loud, then flip

What is a denominator?

The bottom of a fraction.

Which inputs do tan x and sec x skip?

x = + nπ (n any integer), where cos x = 0.

Which inputs do cot x and csc x skip?

x = nπ, where sin x = 0.

Is x = − allowed in cot x?

Yes: − + 4π = , where sin x = 1, not 0, so cot(−) = = 0.

sec = ?

Undefined: − 4π = , where cos x = 0.

True or false: cot x is undefined at every quadrantal angle.

False: it fails only where sin x = 0
cot $π over 2$ = $0 over 1$ = 0.