Signs and Domains of the Trigonometric Functions

14 min read

Type $\sin(100)$ into a calculator and it answers. Type $\tan(\pi/2)$ and it throws an error. That error is honest: the tangent of $\pi/2$ has no value at all. A point on the unit circle has a left-right coordinate $x$ and an up-down coordinate $y$, and those two signs decide where every trig function is positive, where it is negative, and where it simply breaks. There is nothing else to memorize.

By the end of this page you can:

• State the sign of each of the six trig functions in any quadrant.
• Trace each sign back to the sign of $x = \cos t$ or $y = \sin t$ instead of a mnemonic.
• State the domain of each function and name exactly where it is undefined.
• Use a known sign and a quadrant to pin down a second function’s sign.

The floor. On the unit circle the terminal point of $t$ is $(x, y) = (\cos t, \sin t)$. Cosine carries the sign of $x$, sine carries the sign of $y$, and any function built as a quotient is undefined wherever its denominator coordinate is $0$. Those three sentences force every fact below.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet points to a quick refresher, not a grade.

• I can state that the terminal point of a real number $t$ on the unit circle has coordinates $(x, y) = (\cos t, \sin t)$. (The Unit Circle: Sine and Cosine)
• I can write the four quotient and reciprocal functions in terms of $x$ and $y$: $\tan t = y/x$, $\cot t = x/y$, $\sec t = 1/x$, $\csc t = 1/y$.
• I can name which half-planes carry negative $x$ and negative $y$ (left half and bottom half).

A 60-second self-check. Pick an answer, then reveal the reasoning.

The Idea: Read the Signs Off the Coordinates

Each of the six trig functions is built from the two coordinates of the terminal point. Cosine returns $x$, sine returns $y$, and the other four are quotients or reciprocals of those two numbers. So the sign of any trig function is just the sign of an expression in $x$ and $y$, and that depends only on which half-planes $x$ and $y$ fall in.

The two axes split the plane into four quadrants. Crossing the vertical axis flips the sign of $x$, so cosine and its reciprocal secant flip there. Crossing the horizontal axis flips the sign of $y$, so sine and its reciprocal cosecant flip there. Tangent and cotangent use both coordinates, so they are positive exactly when $x$ and $y$ agree in sign.

Domains come from the same coordinates, now asking about division by zero. On the unit circle $x = 0$ on the vertical axis (top and bottom) and $y = 0$ on the horizontal axis (left and right). A function with $x$ in a denominator (tangent and secant) is undefined where $x = 0$. A function with $y$ in a denominator (cotangent and cosecant) is undefined where $y = 0$. Sine and cosine have no denominator, so they are defined for every real number.

Try it. The terminal point of some $t$ lands in Quadrant II, where $x < 0$ and $y > 0$. Decide the sign of $\cos t$, $\sin t$, and $\tan t$ with no chart.

Prerequisite Hub

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Builds on:

Unlocks (what this section feeds into next):

Cross-course prerequisites: none.

Official Definitions

The six functions are defined from the terminal point $(x, y)$ of a real number $t$ on the unit circle. Stewart 7e, Section 5.2 states them this way.

The six functions in coordinate form

Let $t$ be any real number and let $P(x, y)$ be the terminal point on the unit circle determined by $t$. Then $\sin t = y$, $\cos t = x$, $\tan t = \dfrac{y}{x}\ (x \neq 0)$, $\csc t = \dfrac{1}{y}\ (y \neq 0)$, $\sec t = \dfrac{1}{x}\ (x \neq 0)$, $\cot t = \dfrac{x}{y}\ (y \neq 0)$.

Stewart 7e, Section 5.2 Trigonometric Functions of Real Numbers.

The conditions $x \neq 0$ and $y \neq 0$ printed beside four of the definitions are the domain restrictions, stated at the source.

Sign of a trig function

The sign of a trig function at $t$ is the algebraic sign (positive or negative) of its value. Because each function is an expression in $x$ and $y$, its sign is the sign of that expression at the terminal point.

Undefined

A function is undefined at an input when no real-number output is assigned to it. A quotient with denominator $0$ is undefined. Here $\tan t$ and $\sec t$ are undefined when $x = 0$, and $\cot t$ and $\csc t$ are undefined when $y = 0$.

OpenStax presents the same restrictions through the reciprocal functions: secant is the reciprocal of cosine, cosecant is the reciprocal of sine, and cotangent is the reciprocal of tangent, so each is undefined exactly where the function it inverts equals zero.

OpenStax Precalculus 2e, Section 5.3 The Other Trigonometric Functions (open alternate).

The Sign Chart and Where It Comes From

The signs of $x$ and $y$ in each quadrant are fixed by which half-planes the quadrant occupies.

From those two columns, the sign of every function follows by substitution. Tangent is $y/x$, so it is positive when $x$ and $y$ share a sign (Quadrants I and III) and negative when they differ (Quadrants II and IV). A reciprocal keeps the sign of what it inverts.

Reading the table by “which functions are positive” gives the familiar pattern: all six positive in I, only sine and cosecant in II, only tangent and cotangent in III, only cosine and secant in IV. That pattern is a consequence of the coordinate signs, not a separate rule. The advantage of grounding it in $x$ and $y$ is that the table can be rebuilt from the two half-plane facts whenever the mnemonic slips.

Domains: Where Each Function Is Undefined

On the unit circle, $x = 0$ at the top and bottom points, the real numbers $t = \pi/2$, $t = 3\pi/2$, and every odd multiple of $\pi/2$. The coordinate $y = 0$ at the right and left points, the real numbers $t = 0$, $t = \pi$, and every integer multiple of $\pi$.

Here $n$ ranges over the integers. The pairing is the content: the two functions with $x$ downstairs (tangent and secant) fail together on the vertical axis, and the two functions with $y$ downstairs (cotangent and cosecant) fail together on the horizontal axis.

Worked Examples

Worked Example 1: Signs of all six functions in a quadrant

Problem. The terminal point of $t$ is in Quadrant III. State the sign of all six trig functions of $t$.

Predict first. Quadrant III is the bottom-left, both coordinates negative. A function using one coordinate should be negative; a function using both (a quotient of two negatives) should be positive.

Step 1. Record the coordinate signs in Quadrant III: $x < 0$ and $y < 0$.

Step 2. Substitute into each function.

\[ \cos t = x < 0, \qquad \sec t = \tfrac{1}{x} < 0. \] \[ \sin t = y < 0, \qquad \csc t = \tfrac{1}{y} < 0. \] \[ \tan t = \tfrac{y}{x} = \tfrac{(-)}{(-)} > 0, \qquad \cot t = \tfrac{x}{y} = \tfrac{(-)}{(-)} > 0. \]

Step 3. Collect the verdict: $\sin t$, $\cos t$, $\sec t$, $\csc t$ are negative; $\tan t$ and $\cot t$ are positive.

Check the prediction. The two single-coordinate functions came out negative and the two quotient functions positive, matching the prediction. This is the standard “tangent and cotangent positive in Quadrant III” fact, now derived rather than recited.

Worked Example 2: A specific input on an axis

Problem. Evaluate the six functions, where defined, at $t = \pi/2$, whose terminal point is $(0, 1)$.

Predict first. With $x = 0$, any function dividing by $x$ should fail, so tangent and secant are the likely casualties.

Step 1. Read the coordinates: $x = 0$, $y = 1$.

Step 2. Evaluate the functions with no $x$ in the denominator. \[ \sin\tfrac{\pi}{2} = y = 1, \qquad \cos\tfrac{\pi}{2} = x = 0, \qquad \csc\tfrac{\pi}{2} = \tfrac{1}{y} = 1, \qquad \cot\tfrac{\pi}{2} = \tfrac{x}{y} = \tfrac{0}{1} = 0. \]

Step 3. Test the functions that divide by $x = 0$. \[ \tan\tfrac{\pi}{2} = \tfrac{y}{x} = \tfrac{1}{0} \quad\text{undefined}, \qquad \sec\tfrac{\pi}{2} = \tfrac{1}{x} = \tfrac{1}{0} \quad\text{undefined}. \]

Check the prediction. Tangent and secant are exactly the two undefined functions. Notice $\cot(\pi/2) = 0$ is fully defined, built from the same numbers; the difference is that $0$ sits in its numerator, not its denominator. Division into zero is fine; division by zero is not.

Worked Example 3: Pin down a sign from partial information

Problem. Suppose $\sin t < 0$ and $\tan t > 0$. In which quadrant does the terminal point of $t$ lie, and what is the sign of $\cos t$?

Predict first. A negative sine points to the bottom half (Quadrants III and IV). A positive tangent needs $x$ and $y$ to agree in sign. Combining them should single out one quadrant.

Step 1. $\sin t = y < 0$ places the point in the bottom half: Quadrant III or IV.

Step 2. $\tan t = y/x > 0$ requires $x$ and $y$ to share a sign. Since $y < 0$, this forces $x < 0$.

Step 3. A point with $x < 0$ and $y < 0$ is in Quadrant III. Because $\cos t = x < 0$, the cosine is negative.

Check the prediction. The point is in Quadrant III with $\cos t < 0$, the single quadrant where sine is negative and tangent is positive at once. This is the reasoning the fundamental-identities section reuses when a Pythagorean identity gives a magnitude and the quadrant must choose the sign.

Worked Example 4: State a domain in set form

Problem. State the domain of $\cot t$ and explain which points on the unit circle are removed.

Predict first. Cotangent is $x/y$, so $y$ is the denominator. The points with $y = 0$ are the left and right ends of the horizontal axis, so those inputs should be removed.

Step 1. Write the function as a quotient: $\cot t = x/y$.

Step 2. Find where the denominator $y = 0$. The terminal point has $y = 0$ at $(1, 0)$ and $(-1, 0)$, which are $t = 0$ and $t = \pi$, and every input differing from these by a full multiple of $\pi$: $t = n\pi$ for an integer $n$.

Step 3. State the domain: all real numbers $t$ except $t = n\pi$ for integer $n$.

Check the prediction. The removed inputs are exactly the right and left ends of the horizontal axis. Secant, by contrast, is removed at the top and bottom, because secant divides by $x$ and cotangent divides by $y$.

Common Misconceptions

Practice Problems

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it hold.

Go Deeper (optional)

None of this is needed to state a sign or a domain or to pass the section. It is here for readers who want to see where the idea leads.

The undefined points become the asymptotes of the graphs. When tangent, cotangent, secant, and cosecant are plotted, each input where the function is undefined turns into a vertical asymptote, a line the curve runs toward without touching. The tangent graph has vertical asymptotes spaced $\pi$ apart, at $t = \pi/2 + n\pi$, for exactly the domain reason derived above: those are the inputs where $x = 0$. The graphs section builds on this directly.

Why a careers reader might care. Any field that samples a periodic signal has to know where a model output is undefined before feeding it to a machine. A control system computing a heading from $\arctan(y/x)$ must guard the case $x = 0$, or the computation divides by zero and the program crashes. Knowing in advance that the bad inputs are precisely the points on one axis lets an engineer write the guard correctly the first time. The same question, where does the denominator vanish, carries into calculus, where it marks the points a function cannot be evaluated and often the points where its behavior is most interesting.

Mastery Checklist

Novice (Level 1-2):

• I can state the sign of $\sin t$, $\cos t$, and $\tan t$ in any quadrant from the signs of $x$ and $y$.
• I can name the two quadrants where $x$ is negative and the two where $y$ is negative.
• I can say which functions are positive in each quadrant without leaning on a mnemonic.

Competent (Level 3-4):

• I can evaluate the six functions at an axis point and name which ones are undefined there.
• I can recover a quadrant from two given signs and read off a third function’s sign.
• I can pair each undefined function with the coordinate in its denominator.

Proficient (Level 5):

• I can state the domain of each of the six functions as a set of real numbers with the excluded inputs named.
• I can justify, with a specific value of $t$, why secant and cosecant are undefined at different inputs.
• I can explain why dividing into zero is allowed but dividing by zero is not, using $\cot(\pi/2) = 0$ versus $\tan(\pi/2)$ undefined.

Connections

Looking back:

• The Unit Circle: Sine and Cosine defines the terminal point $(x, y) = (\cos t, \sin t)$ that every sign and domain on this page reads from.
• Reference Angles gives the size of a trig value; the quadrant gives its sign, and the two together produce the exact value.

Looking ahead:

• Fundamental Identities: a Pythagorean identity often gives a value up to its sign, and the quadrant decides which sign to take.
• Graphs of the Other Trigonometric Functions: the inputs where tangent, cotangent, secant, and cosecant are undefined become the vertical asymptotes of their graphs.

Real-world connections:

• A program that computes an angle from coordinates with a quotient must guard the inputs where a denominator coordinate is zero, the same inputs that make tangent or cotangent undefined.

Resources

Last updated: 2026-06-24