Cotangent: the branch that falls

Explain it like I am five

Picture a playground slide squeezed between two walls. Near the left wall it starts higher than any ladder; it slopes down, passes ground level exactly halfway, and near the right wall it drops deeper than any pit. That is one branch (one unbroken piece) of cotangent, cot x = $\frac{\cos x}{\sin x}$ , cosine divided by sine.

The walls stand where the bottom, sin x, is 0: at x = 0, at x = π, and at every multiple nπ (n any integer). Between them sine is positive while cosine slides from 1 to −1, so the fraction falls. Three points, with $square root of 2 over 2$ (root 2 over 2, about 0.71):
at $π over 4$ , $square root of 2 over 2$ ÷ $square root of 2 over 2$ = 1;
at $π over 2$ , 0 ÷ 1 = 0;
at $3 π over 4$ , − $square root of 2 over 2$ ÷ $square root of 2 over 2$ = −1.

So the branch passes ( $π over 4$ , 1), ( $π over 2$ , 0) and ( $3 π over 4$ , −1), falling from left to right. Decreasing means exactly this: a bigger input gives a smaller output. The same branch repeats every π.

In plain words

Picture a road that descends between two closed gates. Near the first gate it is higher than any height you name. As you walk right, it drops through height 1, then 0, then −1, and keeps dropping near the second gate. That is one cotangent branch. Decreasing means a larger input gives a smaller output within that branch. The gates sit at 0 and π radians because those inputs make sine zero. Inside the gates, cotangent is continuous. At the midpoint $π over 2$ , it has the ordinary output 0, even though tangent has no output there.

The highlighted radian inputs give falling heights; the two endpoint columns are excluded.

Reminder

Coordinates and plotting. Put input first: cot( $3 π over 4$ ) = −1 gives the point ( $3 π over 4$ , −1).
Radian fractions. To copy left, subtract π with a common denominator: $π over 4$ − $4 π over 4$ = − $3 π over 4$ .
Integer multiples. x = nπ includes −π, 0, and π by taking n = −1, 0, and 1.
Unit-circle coordinates. At $π over 2$ the circle point is (0, 1), so cosine is 0 and sine is 1; cotangent uses 0 ÷ 1.
Fraction division and signs. A negative divided by a positive is negative: − $square root of 2 over 2$ ÷ $square root of 2 over 2$ = −1.
Zero numerator and denominator. $0 over 1$ = 0 is allowed, but $1 over 0$ is undefined. Check the denominator before dividing.
Inequalities and interval endpoints. (0, π) excludes both endpoints; (0, $π over 2$ ] excludes 0 and includes $π over 2$ .
Substitution checks. After locating a boundary, put it back into the denominator: sin π = 0 confirms that π must be excluded.
Odd symmetry. A half-turn changes both coordinates: ( $π over 4$ , 1) pairs with (− $π over 4$ , −1).
Quadrant signs. Quadrant I is upper right, so cos ÷ sin is positive. Quadrant II is upper left, so negative cos ÷ positive sin is negative.

Why it works. Cotangent divides cosine by sine. On 0 < x < π, sine is positive. Cosine changes from positive to 0 to negative, so cotangent passes from positive through 0 to negative. The unit-circle ray turns from almost horizontal right to vertical to almost horizontal left; horizontal coordinate divided by positive vertical coordinate decreases throughout. Near 0 the quotient grows positively without bound; near π it grows negatively without bound. Adding π changes both signs and leaves their quotient unchanged.

RuleRule: cot x = , with x ≠ nπ for every integer n. Its vertical asymptotes are x = nπ and its period is π.
On (0, π), the continuous decreasing branch passes through (, 1), (, 0), and (, −1).

The same idea, five ways

Say it

Say: cotangent decreases within each branch, so farther right gives a lower output before the next asymptote.

Write it

Write: the cotangent branch on (0, π) falls continuously from unbounded positive heights to unbounded negative heights.

In math

cot x = $\frac{\cos x}{\sin x}$
0 < x < π
(0, π)
{x | 0 < x < π}
If 0 < $x sub 1$ < $x sub 2$ < π, then cot $x sub 1$ > cot $x sub 2$ .
cot( $π over 2$ ) = 0
x = nπ: vertical asymptotes
Graph words: separate falling branches

Walk down a road between two closed gates; every move right within this stretch puts you lower.

See it

With x in radians, the branch falls through the three marked points between 0 and π.

The same idea, other ways

As a falling road

Begin immediately right of 0, where the branch is high and positive. Move right through the point of height 1, cross the horizontal axis at $π over 2$ , and continue through height −1. The curve stays continuous inside the interval and drops without bound near π from the left.

The radian branch travels from positive heights through zero to negative heights.

As a unit-circle ratio

Cotangent is horizontal coordinate divided by vertical coordinate. At $π over 2$ the point is (0, 1), so the quotient is 0 ÷ 1 = 0. Tangent reverses that order and would divide 1 by 0. The two formulas do not have the same domain.

At radians, cotangent divides the horizontal coordinate 0 by the vertical coordinate 1.

As three columns

The output under $π over 4$ is 1, the output under $π over 2$ is 0, and the output under $3 π over 4$ is −1. Read the highlighted columns from left to right: their heights fall. The graph extends beyond those three sample heights without a highest or lowest output, so its range is all real numbers.

Increasing radian inputs on this branch give the three decreasing output heights.

x in radians	cot x
0	undefined
$π over 4$	1
$π over 2$	0
$3 π over 4$	−1
π	undefined

Strategy: step by step

1. Find where sin x = 0. This finds the inputs that make cotangent's denominator zero so those inputs can be excluded.
2. For one branch, draw dashed lines at x = 0 and x = π. Plug the endpoints back into sine: sin 0 = 0 and sin π = 0.
3. Read the three table columns to plot ( $π over 4$ , 1), ( $π over 2$ , 0), and ( $3 π over 4$ , −1).
4. Draw one smooth falling branch through the points. Near 0 from the right it rises without bound; near π from the left it falls without bound.
5. Copy the branch every π units left or right. Keep each copied branch separate at x = nπ.

Strategy

Strategy: sketch cotangent from sine zeros and a falling branch

Is sin x = 0 at the requested input?

YesThe cotangent output is undefined; use the input as a vertical asymptote.

NoUse cosine divided by sine to evaluate the accepted input.

↓

Is tangent undefined at this input?

YesStill use cosine divided by sine. Cotangent can be defined even when tangent is undefined.

NoThe quotient formula remains valid; do not replace its domain check.

↓

Does the question ask for more than the branch on (0, π)?

YesCopy the falling branch every π and retain separate asymptotes.

NoDraw only the continuous branch between 0 and π.

Locate and check the sine zeros to draw boundary lines at consecutive multiples of π.
Mark the zero halfway between the boundaries.
Mark height 1 one quarter-period after the left boundary and height −1 one quarter-period before the right boundary.
Draw the falling continuous branch, then copy it every π.

Worked exampleThe cotangent branch between π and 2π

Graph the branch of y = cot x that passes through the point ( $3 π over 2$ , 0), then extend the graph using the period. The three-column table gives exact values on this branch: cot( $4 π over 3$ ) = $square root of 3 over 3$ , cot( $3 π over 2$ ) = 0 and cot( $5 π over 3$ ) = − $square root of 3 over 3$ . The table lists each input as a count of sixths of π, so 8, 9 and 10 stand for $8 π over 6$ , $9 π over 6$ and $10 π over 6$ . It rounds the outputs to two decimal places: 0.58, 0 and −0.58.

Table of y = cot x on the branch between π and 2π. The inputs are 8, 9 and 10 sixths of π, which are , and . The outputs 0.58, 0 and −0.58 are the rounded values of , 0 and −.

Solve sin x = 0. Sine is zero at x = …, −π, 0, π, 2π, 3π, …, which is x = nπ for every integer n. Exclude these inputs, so the domain of y = cot x is x ≠ nπ.cot x = $\frac{\cos x}{\sin x}$ has sin x in the denominator, and division by 0 is undefined. At x = nπ the numerator cos x is 1 or −1, not 0. So the size of the quotient grows without bound near x = nπ, which makes each line x = nπ a vertical asymptote.
Since π < $3 π over 2$ < 2π, the branch through ( $3 π over 2$ , 0) lies between π and 2π, two neighboring zeros of sine. Draw dashed vertical lines at x = π and x = 2π. Plug the endpoints back into sine: sin π = 0 and sin 2π = 0.A cotangent branch runs from one asymptote to the next. No multiple of π lies strictly between π and 2π, so the curve is unbroken there. Checking sine at both endpoints confirms that the dashed lines sit exactly where the denominator is zero.
Read the three table columns. Convert each input: $8 π over 6$ = $4 π over 3$ , $9 π over 6$ = $3 π over 2$ and $10 π over 6$ = $5 π over 3$ . Pair them with the exact outputs $square root of 3 over 3$ , 0 and − $square root of 3 over 3$ , which the table shows as 0.58, 0 and −0.58. Plot ( $4 π over 3$ , $square root of 3 over 3$ ), ( $3 π over 2$ , 0) and ( $5 π over 3$ , − $square root of 3 over 3$ ).The table counts sixths of π so that every input is a whole number. Dividing a count by 6 and simplifying turns it back into an angle. All three inputs lie between π and 2π. The zero sits exactly halfway between the asymptotes, because (π + 2π) ÷ 2 = $3 π over 2$ .
Draw one smooth curve through the three points that falls from left to right. On the right side of x = π the curve rises without bound along the dashed line. On the left side of x = 2π it falls without bound.On the interval (π, 2π), sin x is negative. Just right of π, cos x is near −1 and sin x is a tiny negative number, so $\frac{\cos x}{\sin x}$ is a large positive number. Just left of 2π, cos x is near 1 and sin x is a tiny negative number, so the quotient is a large negative number. The table agrees: as x increases, the outputs go $square root of 3 over 3$ , then 0, then − $square root of 3 over 3$ .
Copy the branch every π units. Moving the three points π to the left gives ( $π over 3$ , $square root of 3 over 3$ ), ( $π over 2$ , 0) and ( $2 π over 3$ , − $square root of 3 over 3$ ) between x = 0 and x = π. Moving them π to the right gives ( $7 π over 3$ , $square root of 3 over 3$ ), ( $5 π over 2$ , 0) and ( $8 π over 3$ , − $square root of 3 over 3$ ) between x = 2π and x = 3π. Draw a dashed asymptote at every x = nπ and keep each copy separate there.Cotangent has period π, so cot(x + π) = cot x. Every branch is therefore the same falling curve shifted by a multiple of π. The function is undefined at each x = nπ, so no copy is joined to its neighbor across an asymptote.

Answer

The branch lies between the dashed asymptotes x = π and x = 2π. It passes through (, ), (, 0) and (, −) and falls from left to right. It rises without bound as x approaches π from the right and falls without bound as x approaches 2π from the left. The full graph repeats this branch between x = nπ and x = (n + 1)π for every integer n, with a dashed asymptote at each x = nπ.

Check Compute the table values from cot x = . First, cos() = − and sin() = −, so cot() = (−) ÷ (−) = = ≈ 0.58. Next, cos() = 0 and sin() = −1, so cot() = 0. Last, cos() = and sin() = −, so cot() = − ≈ −0.58. The left copy also agrees with the rule's branch on (0, π). The point (, ) falls between (, 1) and (, 0), and (, −) falls between (, 0) and (, −1).

Work to write

sin x = 0 at x = nπ, so cot x is undefined there: x ≠ nπ
Asymptotes x = π and x = 2π; sin π = 0 and sin 2π = 0
$8 π over 6$ = $4 π over 3$ , $9 π over 6$ = $3 π over 2$ , $10 π over 6$ = $5 π over 3$
Plot ( $4 π over 3$ , $square root of 3 over 3$ ), ( $3 π over 2$ , 0), ( $5 π over 3$ , − $square root of 3 over 3$ )
Falling branch: rises without bound just right of x = π, falls without bound just left of x = 2π
Period π: copy the branch between each x = nπ and x = (n + 1)π, kept separate at x = nπ

The branch lies between the dashed asymptotes x = π and x = 2π. It passes through ( $4 π over 3$ , $square root of 3 over 3$ ), ( $3 π over 2$ , 0) and ( $5 π over 3$ , − $square root of 3 over 3$ ) and falls from left to right. It rises without bound as x approaches π from the right and falls without bound as x approaches 2π from the left. The full graph repeats this branch between x = nπ and x = (n + 1)π for every integer n, with a dashed asymptote at each x = nπ.

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: evaluate the midpoint zero

Find cot( $π over 2$ ). This asks for the output at the accepted input $π over 2$ , so use cotangent's own denominator rather than tangent's domain.

The point at radians verifies cotangent's allowed zero output.

Read cos( $π over 2$ ) = 0 and sin( $π over 2$ ) = 1 from the unit-circle point (0, 1).The first coordinate is cosine, and the second is sine.
Check that the denominator 1 is nonzero, then calculate $0 over 1$ = 0.Division is allowed when the denominator is nonzero, including when the numerator is 0.

Answer

cot() = 0.

Check The unit-circle point on the vertical ray at has horizontal coordinate 0 and positive vertical coordinate 1. Their cotangent ratio is therefore 0.

Rung 2Rung 2: draw the positive half of the branch

Sketch cotangent for 0 < x ≤ $π over 2$ , including its point at $π over 4$ . This asks for the positive half of a branch and the midpoint zero, while excluding 0.

The radian input 0 is excluded, while the highlighted columns provide two allowed points.

The positive half-branch falls from the asymptote to its included zero at radians.

Exclude 0 because sin 0 = 0; draw the dashed line x = 0.Substitution finds and verifies the zero denominator at the left boundary.
At $π over 4$ , divide $square root of 2 over 2$ by itself to get 1, then plot ( $π over 4$ , 1).Equal nonzero cosine and sine values give the quotient 1.
Include ( $π over 2$ , 0) and draw a falling curve toward it from high positive heights near 0.The denominator is positive inside this half-interval, and the quotient decreases toward its allowed zero.

Answer

Excluded boundary: x = 0.
Points: ( $π over 4$ , 1), ( $π over 2$ , 0).
Shape: positive and falling on (0, $π over 2$ ), with the zero at $π over 2$ included.

Check Quadrant I has positive sine and cosine, so cotangent must stay positive before ; the axis point at supplies the included zero.

Rung 3Rung 3: complete the falling branch

Complete the cotangent sketch on (0, π). This asks you to add the negative half and explain why π, like 0, is excluded.

The last highlighted radian column supplies the point that completes the branch.

The complete radian branch falls through all three points toward its right asymptote.

At $3 π over 4$ , divide − $square root of 2 over 2$ by $square root of 2 over 2$ to get −1.Cosine is negative and sine positive, with equal nonzero magnitudes.
Plot ( $3 π over 4$ , −1) after the midpoint zero and continue the smooth falling curve.The output row falls from 1 to 0 to −1 as the radian inputs increase.
Check sin π = 0 and draw x = π as the right vertical asymptote.This denominator check excludes the endpoint rather than assigning it a value.
Extend the curve negatively without bound near π from the left.Cosine is near −1 while sine is tiny and positive, so their quotient is very negative.

Answer

Points: ( $π over 4$ , 1), ( $π over 2$ , 0), ( $3 π over 4$ , −1).
Boundary lines: x = 0, x = π.
Shape: one continuous falling branch with unbounded ends.

Check In quadrant II cosine is negative and sine positive, so the added half must be negative. That independent sign check agrees with the point at and the downward right end.

Rung 4Rung 4: repeat left and check origin symmetry

Sketch the cotangent branch on (−π, 0) and check its points using odd symmetry. This asks you to copy the known branch one period left, then verify it by a half-turn about the origin.

The negative radian inputs retain the falling output pattern 1, 0, −1.

The radian branches repeat by π and pair under a half-turn about the origin.

Subtract π from the boundaries 0 and π to get −π and 0.Period π preserves the branch's output pattern one π unit left.
Subtract π = $4 π over 4$ from the key inputs: $π over 4$ − π = − $3 π over 4$ , $π over 2$ − π = − $π over 2$ , and $3 π over 4$ − π = − $π over 4$ .Common denominators let you subtract the π fractions exactly.
Keep the corresponding outputs 1, 0, and −1, and draw one falling branch.The copied input columns retain the original output row.
Check sin(−π) = 0 and sin 0 = 0, so both new boundary inputs remain excluded.Plugging the boundaries into cotangent's denominator confirms the two vertical asymptotes.
Rotate ( $π over 4$ , 1) to (− $π over 4$ , −1), and ( $3 π over 4$ , −1) to (− $3 π over 4$ , 1).Odd symmetry changes the signs of both coordinates and checks the copied points a second way.

Answer

Interval: (−π, 0).
Boundary lines: x = −π, x = 0.
Points: (− $3 π over 4$ , 1), (− $π over 2$ , 0), (− $π over 4$ , −1).
Shape: a continuous falling branch, matching origin symmetry.

Check Use the quotient directly at : cosine is positive and sine negative with equal magnitudes, so cotangent is −1. At −, both cosine and sine are −, so their quotient is 1, agreeing with the copied and rotated points.

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

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

Cotangent's quotient has sine in the denominator, and sine at equals 1. The expression 1 divided by tangent is unavailable when tangent itself is undefined.

✓ Instead: Use cosine divided by sine: cot() = = 0. The reciprocal identity cot x = requires both sides to be defined.

✗ Not this: Counterexample: draw a rising cotangent branch through (, 1), (, 0), and (, −1).

Those outputs decrease from 1 to 0 to −1 as the inputs increase. A rising curve contradicts its own plotted points.

✓ Instead: Draw a falling branch between consecutive multiples of π.

Tips and tricks

Tip: cotangent falls, and its zero is midway between consecutive multiples of π.
Tip: use cotangent's own quotient cos x divided by sin x when checking whether an input is allowed.
Tip: the whole branch reaches every real output, so cotangent has no highest or lowest height and no amplitude.

Trap. Using cot x = 1 divided by tan x at an input where tangent is undefined. Cotangent's reliable definition is cos x divided by sin x; at that gives the valid output 0.

Keep in mind

Cotangent's walls are sine's zeros, x = nπ, not cosine's: cot x is fine at $π over 2$ , where it equals 0.
Do not compute cot x as 1 ÷ tan x where tan x is undefined; $\frac{\cos x}{\sin x}$ gives cot $π over 2$ = $0 over 1$ = 0 directly.
Near the left wall the heights are huge and positive, near the right wall huge and negative: cot 0.1 ≈ 9.97 and cot 3 ≈ −7.02.

Memory hookTangent climbs, cotangent slides: tan rises between its walls, while cot falls between walls at 0, π, 2π, crossing zero halfway.

Flash cards: say the answer out loud, then flip

Where are cotangent's vertical asymptotes?

x = nπ, where sin x = 0.

Does a cotangent branch rise or fall from left to right?

It falls: cotangent is decreasing on each branch.

Is cot 2.8 positive or negative?

Negative: 2.8 is between ≈ 1.57 and π ≈ 3.14, where cosine is negative and sine positive.

cot = ?

−1: − π = , and the branch passes (, −1).

True or false: cot is undefined because tan is.

False: sin = 1, not 0, so cot = = 0.