Trigonometric Functions of Real Numbers

17 min read

Tape a marker to the edge of a bike wheel and spin it slowly. Track only its height off the ground. That single number rises, falls, and repeats, tracing a smooth wave. The wave is the sine function, and the across-position is the cosine. The six trig functions all come from watching one point go around one circle.

By the end of this page you can:

• State the definition of all six trigonometric functions of a real number $t$ from the terminal point $(x, y)$ on the unit circle.
• Read $\cos t = x$ and $\sin t = y$ directly off a labeled terminal point, and build $\tan t$, $\cot t$, $\sec t$, $\csc t$ as ratios of $x$ and $y$.
• Name which functions are undefined at a terminal point on an axis, and say why.
• Use $\cos^2 t + \sin^2 t = 1$ to confirm a point is on the unit circle.

The floor. One move carries the rest. Find the terminal point $(x, y)$ that the input $t$ lands on. Then $\cos t$ is the across-value $x$, $\sin t$ is the up-value $y$, and the other four functions are ratios of those two coordinates. Find the point first, then read its coordinates.

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 find the terminal point $(x, y)$ on the unit circle for the common inputs such as $\dfrac{\pi}{6}$, $\dfrac{\pi}{4}$, $\dfrac{\pi}{3}$, and the quadrantal inputs $0$, $\dfrac{\pi}{2}$, $\pi$, $\dfrac{3\pi}{2}$. (Terminal Points and the Reference Number)
• I can read the $x$-value and the $y$-value of a labeled point in the coordinate plane and name its quadrant.
• I can read $f(t)$ as one output of a function applied to one input $t$, not as a product. (Function Notation Review)
• I can recognize the unit-circle equation $x^2 + y^2 = 1$.

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

The Idea

Draw a circle of radius $1$ centered at the origin. Start at $(1, 0)$ on the right side and walk counterclockwise along the edge for a distance of exactly $1$ unit. The input $t$ is that distance walked, positive for counterclockwise and negative for clockwise. Where you stop is the terminal point, with an across-value $x$ and an up-value $y$.

Two of the functions read those coordinates straight off. The cosine reports the across-value, $\cos t = x$. The sine reports the up-value, $\sin t = y$. The other four are ratios and reciprocals of the same two numbers:

• The tangent is up over across, $\dfrac{y}{x}$.
• The cotangent is across over up, $\dfrac{x}{y}$.
• The secant is the reciprocal of the across-value, $\dfrac{1}{x}$.
• The cosecant is the reciprocal of the up-value, $\dfrac{1}{y}$.

This is the shift that makes trigonometry a study of functions rather than a study of triangles. In a right triangle the inputs are acute angles and the outputs are ratios of side lengths. On the unit circle the input is any real number and the outputs are coordinates and their ratios. The two views agree on the acute angles, and the circle view is the one that lets the functions accept every real number, not only the angles small enough to fit inside a triangle.

A division can fail. When the across-value $x$ is $0$, the tangent and the secant divide by $0$ and are undefined. When the up-value $y$ is $0$, the cotangent and the cosecant divide by $0$ and are undefined. Reading off two coordinates always works; dividing by one that happens to be zero does not.

Turn the angle and watch the two coordinates move. The teal leg is the across-value (the cosine); the amber leg is the up-value (the sine). Send the point through all four quadrants and watch each value pass through positive and negative.

Prerequisite Hub

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Builds on:

Unlocks (what this section feeds into next):

Cross-course prerequisites: none.

Official Definitions

Unit Circle

A unit circle has center $(0, 0)$ and radius $1$, so its equation is $x^2 + y^2 = 1$. On a unit circle, the length of the intercepted arc equals the radian measure of the central angle $t$.

Source: Stewart 7e, Section 5.1 The Unit Circle. Compare OpenStax Precalculus 2e, Section 5.2.

The Six Trigonometric Functions of a Real Number

Let $t$ be a real number representing the length of an arc on the unit circle with initial point $(1, 0)$ and terminal point $(x, y)$. The six trigonometric functions of $t$ are defined by reading the coordinates of that terminal point:

\[ \sin t = y, \qquad \cos t = x, \qquad \tan t = \frac{y}{x}, \] \[ \csc t = \frac{1}{y}, \qquad \sec t = \frac{1}{x}, \qquad \cot t = \frac{x}{y}. \]

The tangent and secant are not defined when $x = 0$. The cotangent and cosecant are not defined when $y = 0$.

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

The arc length $t$ and the radian measure of the central angle are the same number on a unit circle, because the radius is $1$. That single fact lets the input be a length, an angle in radians, or just a real number, with no contradiction.

Quick Reference

The unit-circle definition. For a real number $t$, let $(x, y)$ be the terminal point of an arc of length $t$ on the unit circle $x^2 + y^2 = 1$. Then

\[ \cos t = x, \quad \sin t = y, \quad \tan t = \frac{y}{x}, \quad \cot t = \frac{x}{y}, \quad \sec t = \frac{1}{x}, \quad \csc t = \frac{1}{y}. \]

The six functions, organized by how each reads the terminal point:

The quadrantal inputs (terminal point on an axis), where two functions are always undefined.

First-quadrant special points (all six values positive here, since $x > 0$ and $y > 0$).

Key Theorem: The Pythagorean Identity

Pythagorean Identity. For any real number $t$, \[ \cos^2 t + \sin^2 t = 1. \]

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

Why this is true. A point $(x, y)$ sits on the unit circle exactly when its distance from the origin is $1$, which means $x^2 + y^2 = 1$. Replace $x$ with $\cos t$ and $y$ with $\sin t$, and the equation becomes $\cos^2 t + \sin^2 t = 1$. The identity is the equation of the circle, written in the language of sine and cosine.

The notation $\cos^2 t$ means $(\cos t)^2$: the cosine of $t$ first, then the square. It does not mean $\cos(t^2)$.

Worked Examples

Example 1: All Six Functions at a Known Terminal Point

The terminal point of an arc of length $t = \dfrac{\pi}{3}$ is $\left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right)$. Find all six trigonometric functions of $\dfrac{\pi}{3}$.

Predict first. The point is in the first quadrant, so all six values should be positive. The across-value is small ($\tfrac{1}{2}$) and the up-value is larger ($\tfrac{\sqrt{3}}{2} \approx 0.87$), so the tangent $\tfrac{y}{x}$ should be larger than $1$.

Compute. Read $x = \dfrac{1}{2}$ and $y = \dfrac{\sqrt{3}}{2}$, then apply each rule.

\[ \cos\frac{\pi}{3} = x = \frac{1}{2}, \qquad \sin\frac{\pi}{3} = y = \frac{\sqrt{3}}{2}. \] \[ \tan\frac{\pi}{3} = \frac{y}{x} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}, \qquad \cot\frac{\pi}{3} = \frac{x}{y} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}. \] \[ \sec\frac{\pi}{3} = \frac{1}{x} = \frac{1}{1/2} = 2, \qquad \csc\frac{\pi}{3} = \frac{1}{y} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}. \]

Compare. All six values are positive, as predicted, and $\tan\dfrac{\pi}{3} = \sqrt{3} \approx 1.73 > 1$, also as predicted. The secant and tangent came out clean because $x = \tfrac{1}{2}$ is not $0$.

Check with the identity. $\cos^2\dfrac{\pi}{3} + \sin^2\dfrac{\pi}{3} = \dfrac{1}{4} + \dfrac{3}{4} = 1$. The terminal point really is on the unit circle.

Example 2: Reading Sine and Cosine off a Known Point

The terminal point of an arc of length $t = \dfrac{\pi}{4}$ is $\left( \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2} \right)$ on the unit circle. Find $\cos\dfrac{\pi}{4}$ and $\sin\dfrac{\pi}{4}$.

Predict first. The point is in the first quadrant and sits on the line $y = x$, halfway up the quarter circle. Both coordinates should be positive and equal, each a bit more than half. Expect two equal positive values near $0.7$.

Compute. The cosine is the $x$-coordinate and the sine is the $y$-coordinate: \[ \cos\frac{\pi}{4} = x = \frac{\sqrt{2}}{2}, \qquad \sin\frac{\pi}{4} = y = \frac{\sqrt{2}}{2}. \]

Compare. Both values are positive and equal, and $\dfrac{\sqrt{2}}{2} \approx 0.707$, which matches the prediction.

Check with the identity. $\cos^2\dfrac{\pi}{4} + \sin^2\dfrac{\pi}{4} = \dfrac{2}{4} + \dfrac{2}{4} = 1$. The point really is on the unit circle.

Cosine is $x$ and sine is $y$, in that order. A terminal point is written $(x, y) = (\cos t, \sin t)$. Reading sine as the first coordinate is one of the most common unit-circle errors. Cosine comes first because $x$ comes first.

Example 3: Finding the Missing Coordinate with the Pythagorean Identity

The terminal point of an arc of length $t$ lands in the second quadrant, and its $x$-coordinate is $-\dfrac{3}{5}$. Find $\sin t$, and then $\tan t$.

Predict first. In the second quadrant the $x$-coordinate is negative and the $y$-coordinate is positive, so $\sin t$ should come out positive and $\tan t = \tfrac{y}{x}$ should come out negative (positive over negative).

Compute the sine. Here $\cos t = -\dfrac{3}{5}$. Substitute into the Pythagorean Identity: \[ \left(-\frac{3}{5}\right)^2 + \sin^2 t = 1 \] \[ \frac{9}{25} + \sin^2 t = 1 \] \[ \sin^2 t = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \] \[ \sin t = \pm\frac{4}{5}. \]

Choose the sign from the quadrant. The terminal point is in the second quadrant, where $y$ is positive, so the sine is positive: \[ \sin t = \frac{4}{5}. \]

Compute the tangent. \[ \tan t = \frac{y}{x} = \frac{4/5}{-3/5} = -\frac{4}{3}. \]

Compare. The sine is positive and the tangent is negative, both as predicted. A quick check: $\left(-\dfrac{3}{5}\right)^2 + \left(\dfrac{4}{5}\right)^2 = \dfrac{9}{25} + \dfrac{16}{25} = 1$.

The identity gives two signs; the quadrant picks one. Taking the square root produces $\pm$, and only the quadrant tells you which sign is correct. State the quadrant before committing to a sign.

Example 4: The Quadrantal Inputs and Where Functions Fail

Find all six functions for $t = 0$ and $t = \dfrac{\pi}{2}$.

Predict first. The terminal point for $t = 0$ is the right side $(1, 0)$, and for $t = \dfrac{\pi}{2}$ it is the top $(0, 1)$. At each one, one coordinate is $0$. Dividing by that zero coordinate will make two of the six functions undefined.

Compute at $t = 0$, terminal point $(1, 0)$. Here $x = 1$, $y = 0$. \[ \cos 0 = 1, \quad \sin 0 = 0, \quad \tan 0 = \frac{0}{1} = 0, \quad \sec 0 = \frac{1}{1} = 1. \] The cotangent $\cot 0 = \dfrac{x}{y} = \dfrac{1}{0}$ and the cosecant $\csc 0 = \dfrac{1}{y} = \dfrac{1}{0}$ divide by $y = 0$, so both are undefined.

Compute at $t = \dfrac{\pi}{2}$, terminal point $(0, 1)$. Here $x = 0$, $y = 1$. \[ \cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1, \quad \cot\frac{\pi}{2} = \frac{0}{1} = 0, \quad \csc\frac{\pi}{2} = \frac{1}{1} = 1. \] The tangent $\tan\dfrac{\pi}{2} = \dfrac{y}{x} = \dfrac{1}{0}$ and the secant $\sec\dfrac{\pi}{2} = \dfrac{1}{x} = \dfrac{1}{0}$ divide by $x = 0$, so both are undefined.

Compare. At each quadrantal input exactly two functions are undefined, and which two depends on which coordinate is zero. When $y = 0$, the cotangent and cosecant fail. When $x = 0$, the tangent and secant fail. That matches the prediction.

Common Misconceptions

Treating $\sin t$ as the product of a symbol named “sin” and the number $t$ (trig-as-algebra-symbols). The expression $\sin t$ is not multiplication. It is the output of the sine function applied to the input $t$, the $y$-coordinate of a specific terminal point. Predict-then-check: if $\sin t$ meant $s \cdot i \cdot n \cdot t$, then $\sin\dfrac{\pi}{2}$ would be some multiple of $\dfrac{\pi}{2} \approx 1.57$, but the actual value is exactly $1$, the height of the top of the circle. No number multiplied by $\dfrac{\pi}{2}$ gives $1$ and also works for every other input. The same warning applies to $\tan t$, $\sec t$, and the rest: each is a single function output, never a product of letters.

Swapping which coordinate is sine and which is cosine (input-output-confusion). The cosine is the first coordinate (horizontal, $x$) and the sine is the second coordinate (vertical, $y$). Predict-then-check at $t = 0$: the terminal point is $(1, 0)$ on the right side. If sine were the $x$-coordinate, then $\sin 0$ would be $1$, but a circle has zero height at its rightmost point, so $\sin 0 = 0$ and $\cos 0 = 1$. Anchor the names to horizontal-versus-vertical, not to the order of the alphabet. The same swap corrupts the ratios: $\tan t = \tfrac{y}{x}$, not $\tfrac{x}{y}$.

Forgetting that the tangent and secant are undefined where $x = 0$ (tan-sec-undefined-where-x-zero). The tangent $\tan t = \tfrac{y}{x}$ and the secant $\sec t = \tfrac{1}{x}$ both have $x$ in the denominator, so both fail wherever the terminal point has across-value $0$, namely at the top $(0,1)$ and the bottom $(0,-1)$. Predict-then-check at $t = \dfrac{\pi}{2}$: the point is $(0, 1)$, so $\tan\dfrac{\pi}{2} = \dfrac{1}{0}$, which is undefined, not $0$. Writing a finite number there is the error. The mirror version applies to $\cot t = \tfrac{x}{y}$ and $\csc t = \tfrac{1}{y}$, which fail where $y = 0$, at the left and right of the circle.

Leveled Practice

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 stick.

Go Deeper (optional)

None of this is needed to evaluate the six functions or to pass the section. It is here for readers who want to see where the unit-circle definition leads.

Why functions of real numbers, not just angles. Tying the input to a length traveled around the circle, rather than to an angle drawn in a triangle, lets sine and cosine accept every real number. That single change is what makes them describe time-varying quantities: a real number $t$ can stand for seconds, and $\sin t$ then traces a value that rises and falls forever. Alternating current, the pressure of a sound wave, and the position of a swinging pendulum are all modeled this way.

Where these functions show up in work. Signal processing, computer graphics, and physics simulations all run on the six functions defined here. A rotation in a graphics engine multiplies coordinates by $\cos t$ and $\sin t$. An audio filter is built from sums of these functions at different rates. The terminal-point definition on this page is the same one those tools compute with.

Mastery Checklist

Novice (Level 1-2):

• I can state that $\cos t = x$ and $\sin t = y$ for the terminal point $(x, y)$ of an arc of length $t$.
• I can write the four ratio rules $\tan t = \tfrac{y}{x}$, $\cot t = \tfrac{x}{y}$, $\sec t = \tfrac{1}{x}$, $\csc t = \tfrac{1}{y}$.
• I can use $\cos^2 t + \sin^2 t = 1$ to check whether a point lies on the unit circle.

Competent (Level 3-4):

• I can find all six functions at a known terminal point, including the special inputs $\tfrac{\pi}{6}$, $\tfrac{\pi}{4}$, $\tfrac{\pi}{3}$.
• I can find a missing coordinate with the Pythagorean Identity and choose the sign from the quadrant.
• I can state the sign of $\cos t$, $\sin t$, and $\tan t$ in each of the four quadrants.

Proficient (Level 5):

• I can name which functions are undefined at each quadrantal input and explain why.
• I can justify the ranges of sine, cosine, secant, and cosecant from both the circle and the identity.
• I can explain why the arc length and the radian measure are the same number on a unit circle.

Connections

Looking back:

• Terminal Points and the Reference Number supplies the terminal point $(x, y)$ that every one of the six functions reads.
• Function Notation Review gives the one-input-one-output meaning that keeps $\sin t$ from being read as a product.

Looking ahead:

• Signs and Domains of Trig Functions settles the sign of each function by quadrant and the exact inputs where each is undefined.
• Fundamental Identities relates the six functions through the reciprocal, Pythagorean, and even-odd identities.
• Graphs of Sine and Cosine plots the outputs against the input to reveal the periodic wave.

Real-world connections:

• A point on a turning wheel traces sine and cosine in its height and its horizontal position.
• Alternating current and sound waves are described by sine and cosine, because each models a quantity that rises and falls smoothly between fixed bounds.

Resources

Last updated: 2026-06-24