Introduction

If right-triangle trigonometry is the vocabulary of trig, the unit circle is the grammar that lets everything else make sense. Recall that with right triangles, sine and cosine are only defined for acute angles: angles strictly between $0°$ and $90°$. The unit circle breaks this restriction entirely. It extends sine, cosine, and tangent to every real-number angle, positive or negative, large or small.

The unit circle is the circle of radius $1$ centered at the origin of the coordinate plane. Its equation is simply:

Every point $(x, y)$ on this circle will turn out to encode a complete trigonometric story.

Radians: A Better Unit

Before diving into the circle itself, we need to talk about how we measure angles. You are likely familiar with degrees, but most serious mathematics uses radians.

One radian is the angle subtended at the center of a circle by an arc whose length equals the radius. Since the circumference of the unit circle is $2\pi$, a full revolution corresponds to exactly $2\pi$ radians. Hence:

To convert between the two systems, let $d$ be an angle in degrees and $r$ its radian measure:

Some conversions worth memorizing:

Degrees	Radians
$30°$	$\dfrac{\pi}{6}$
$45°$	$\dfrac{\pi}{4}$
$60°$	$\dfrac{\pi}{3}$
$90°$	$\dfrac{\pi}{2}$
$180°$	$\pi$
$360°$	$2\pi$

Radians are preferred not merely by convention, it's because they make calculus, series expansions, and virtually every formula in higher mathematics easier. When an angle appears raw in a formula (like $e^{i\theta}$ or $\frac{d}{d\theta}\sin\theta = \cos\theta$), it is always in radians.

Connecting the Circle to Right Triangles

Pick any point $P = (x, y)$ on the unit circle and draw the segment from the origin $O$ to $P$. Drop a perpendicular from $P$ to the $x$-axis, landing at $A = (x, 0)$. This produces a right triangle $OAP$ with:

• Hypotenuse $OP = 1$ (radius of the unit circle)
• Horizontal leg $OA = |x|$
• Vertical leg $AP = |y|$

Credit: Brilliant

Let $\theta$ be the angle at $O$, measured counterclockwise from the positive $x$-axis. Applying the standard SOH-CAH-TOA definitions:

This yields the central fact:

The power of this definition is that it works for any angle. When $\theta$ moves into the second quadrant, $x$ becomes negative, so $\cos\theta < 0$.

The Pythagorean Identity

Since every point on the unit circle satisfies $x^2 + y^2 = 1$, substituting $x = \cos\theta$ and $y = \sin\theta$ gives immediately:

This is the Pythagorean Identity, the single most useful trig identity. Dividing through by $\cos^2\theta$ or $\sin^2\theta$ yields two more:

Signed Angles and the Four Quadrants

An angle on the unit circle is always measured from the positive $x$-axis with the vertex at the origin. It is positive when swept counterclockwise and negative when swept clockwise. The terminal side is the ray from the origin through the point on the unit circle.

Credit: Brilliant

The signs of $\sin\theta$ and $\cos\theta$ depend on which quadrant the terminal side falls in:

Quadrant	$x$	$y$	$\cos\theta$	$\sin\theta$
I ($0 < \theta < \frac{\pi}{2}$)	$+$	$+$	$+$	$+$
II ($\frac{\pi}{2} < \theta < \pi$)	$-$	$+$	$-$	$+$
III ($\pi < \theta < \frac{3\pi}{2}$)	$-$	$-$	$-$	$-$
IV ($\frac{3\pi}{2} < \theta < 2\pi$)	$+$	$-$	$+$	$-$

A useful mnemonic: All Students Take Calculus... going counterclockwise from Quadrant I, the functions that are positive are All, Sine, Tangent, Cosine.

Special Angles

The special angles are those for which the coordinates can be computed exactly from $30$-$60$-$90$ and $45$-$45$-$90$ triangle relationships. Memorizing these is essential for every competition and course involving trigonometry.

Credit: Brilliant

$\theta$ (radians)	$\theta$ (degrees)	$\cos\theta$	$\sin\theta$	$\tan\theta$
$0$	$0°$	$1$	$0$	$0$
$\dfrac{\pi}{6}$	$30°$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\dfrac{1}{\sqrt{3}}$
$\dfrac{\pi}{4}$	$45°$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$
$\dfrac{\pi}{3}$	$60°$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\sqrt{3}$
$\dfrac{\pi}{2}$	$90°$	$0$	$1$	undefined
$\pi$	$180°$	$-1$	$0$	$0$
$\dfrac{3\pi}{2}$	$270°$	$0$	$-1$	undefined

Rather than memorizing this table as a list of facts, notice the pattern in the sine column for the first quadrant:

The numerators run $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ in order. The cosine column runs the same sequence in reverse.

Coordinates in the Unit Circle

A right triangle $AOB$ with right angle at $A$ lies on the Cartesian plane such that $\overline{OA}$ lies on the $x$-axis, point $O$ lies at the origin, and point $B$ lies anywhere on the unit circle. Note that $OB = 1$ unit.

Credit: Brilliant

When defining sine and cosine in terms of the unit circle, lengths can be negative. If $\overline{OA}$ extends along the negative $x$-axis, then $OA$ is negative; if $\overline{AB}$ extends below the $x$-axis, $AB$ is negative. Under this convention:

The sine of an angle is the $y$-coordinate of its point on the unit circle; the cosine is the $x$-coordinate.

Worked Examples

Example 1. A line through the origin meets the unit circle at angle $\theta = \dfrac{\pi}{4}$. Find the coordinates of that point.

Since the coordinates are $(\cos\theta, \sin\theta)$:

So the point is $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$. $\square$

Example 2. A point on the unit circle has $x$-coordinate $\dfrac{1}{2}$. Find $\tan^2\theta$.

Since $x = \cos\theta$, we have $\cos^2\theta = \dfrac{1}{4}$. By the Pythagorean identity:

Therefore:

Allied Angles (Reduction Formulas)

One of the most practical payoffs of the unit circle is a complete set of reduction formulas that let you express any trig function of a "compound" angle like $90° + \theta$ or $270° - \theta$ purely in terms of $\sin\theta$, $\cos\theta$, and $\tan\theta$. Competitions use these constantly; so does every calculus course.

How to derive them?

The key insight is that the unit circle encodes geometry, not algebra. For any angle $\alpha$, the point $(\cos\alpha, \sin\alpha)$ sits on the circle. When you add or subtract a multiple of $90°$, the point rotates to a new quadrant. Reading off the new coordinates tells you exactly which function appears and what sign it carries.

The two-step recipe:

1. Function type : if the shift is an odd multiple of $90°$ (i.e. $90°$, $270°$), the function swaps: $\sin \leftrightarrow \cos$, $\tan \leftrightarrow \cot$, $\sec \leftrightarrow \csc$. If the shift is an even multiple of $90°$ (i.e. $0°$, $180°$, $360°$), the function stays the same.

2. Sign : imagine $\theta$ is a small positive acute angle and ask: in which quadrant does the compound angle land? Use ASTC to determine the sign of that function there.

That's it. You can derive any entry in the tables below on the fly.

Worked derivation: $\sin(90° + \theta)$:

Place $\theta$ in Quadrant I, so the point is $(\cos\theta, \sin\theta)$. Rotating $90°$ counterclockwise maps $(x, y) \mapsto (-y, x)$, so the new point is $(-\sin\theta, \cos\theta)$. The $y$-coordinate of this new point is $\cos\theta$, and $90° + \theta$ lands in Quadrant II where sine is positive. Hence:

Every formula below follows from the same rotation argument.

Shifts by $90°$

Function	$90° - \theta$	$90° + \theta$
$\sin$	$\cos\theta$	$\cos\theta$
$\cos$	$\sin\theta$	$-\sin\theta$
$\tan$	$\cot\theta$	$-\cot\theta$
$\cot$	$\tan\theta$	$-\tan\theta$
$\sec$	$\csc\theta$	$-\csc\theta$
$\csc$	$\sec\theta$	$\sec\theta$

Function swaps (sin↔cos, tan↔cot, sec↔csc). Sign from ASTC: $90° - \theta$ stays in Q I (all positive); $90° + \theta$ lands in Q II (only sine and cosecant positive).

Shifts by $180°$

Function	$180° - \theta$	$180° + \theta$
$\sin$	$\sin\theta$	$-\sin\theta$
$\cos$	$-\cos\theta$	$-\cos\theta$
$\tan$	$-\tan\theta$	$\tan\theta$
$\cot$	$-\cot\theta$	$\cot\theta$
$\sec$	$-\sec\theta$	$-\sec\theta$
$\csc$	$\csc\theta$	$-\csc\theta$

Function stays the same. $180° - \theta$ is in Q II (sin and csc positive, rest negative); $180° + \theta$ is in Q III (tan and cot positive, rest negative).

Shifts by $270°$

Function	$270° - \theta$	$270° + \theta$
$\sin$	$-\cos\theta$	$-\cos\theta$
$\cos$	$-\sin\theta$	$\sin\theta$
$\tan$	$\cot\theta$	$-\cot\theta$
$\cot$	$\tan\theta$	$-\tan\theta$
$\sec$	$-\csc\theta$	$\csc\theta$
$\csc$	$-\sec\theta$	$-\sec\theta$

Function swaps again (odd multiple of $90°$). $270° - \theta$ is in Q III (tan and cot positive); $270° + \theta$ is in Q IV (cos and sec positive).

Shifts by $360°$ (and negative angles)

Function	$-\theta$	$360° - \theta$	$360° + \theta$
$\sin$	$-\sin\theta$	$-\sin\theta$	$\sin\theta$
$\cos$	$\cos\theta$	$\cos\theta$	$\cos\theta$
$\tan$	$-\tan\theta$	$-\tan\theta$	$\tan\theta$
$\cot$	$-\cot\theta$	$-\cot\theta$	$\cot\theta$
$\sec$	$\sec\theta$	$\sec\theta$	$\sec\theta$
$\csc$	$-\csc\theta$	$-\csc\theta$	$\csc\theta$

Function stays the same (even multiple of $90°$). $360° + \theta$ is just periodicity, a full revolution returns to the same point. $-\theta$ and $360° - \theta$ both reflect across the $x$-axis: the $y$-coordinate negates while $x$ stays fixed, which is exactly why sine is an odd function ($\sin(-\theta) = -\sin\theta$) and cosine is an even function ($\cos(-\theta) = \cos\theta$).

Quick-fire examples

Simplify $\tan(180° + \theta)$.

$180°$ shift → function stays the same ($\tan$). $180° + \theta$ lands in Q III where tangent is positive. So $\tan(180° + \theta) = \tan\theta$. $\square$

Simplify $\cos(270° - \theta)$.

$270°$ shift → function swaps ($\cos \to \sin$). $270° - \theta$ lands in Q III where cosine is negative. So $\cos(270° - \theta) = -\sin\theta$. $\square$

Simplify $\csc(90° + \theta)$.

$90°$ shift → function swaps ($\csc \to \sec$). $90° + \theta$ lands in Q II where cosecant is positive. So $\csc(90° + \theta) = \sec\theta$. $\square$

Contest Problems

Unit Circle Sign Problem

If $R = \sin 130° + \cos 130°$, which of the following is true?

$R = 0 \qquad R \leq 0 \qquad R < 0 \qquad R > 0$

Credit: Brilliant

The angle $130°$ lies in Quadrant II, so $\sin 130° > 0$ and $\cos 130° < 0$. To determine the sign of their sum, note that $130° = 180° - 50°$. Using the cofunction shift:

So $R = \sin 50° - \cos 50°$. Since $50° > 45°$, we have $\sin 50° > \cos 50°$ (sine exceeds cosine for angles past $45°$ in the first quadrant), hence:

$\boxed{R > 0}$

The Unit Circle Toolkit

Much like GCD and LCM problems reduce to isolating one prime at a time, unit circle problems reduce to identifying the quadrant and reference angle. Here is a summary of the key tools:

Situation	Tool
Evaluating $\sin\theta$ or $\cos\theta$ for a special angle	Unit circle table
Angle outside $[0°, 360°]$	Reduce modulo $360°$ (or $2\pi$)
Angle in Quadrant II/III/IV	Reference angle + ASTC sign rule
Equation involving $\sin^2\theta + \cos^2\theta$	Pythagorean identity
Expression mixing $\sin$ and $\tan$	Convert to $\sin$/$\cos$, simplify
Degree–radian conversion	$r = \dfrac{\pi d}{180}$

Reference Angles

For any angle $\theta$, the reference angle $\hat\theta$ is the acute angle between the terminal side and the $x$-axis. The absolute values $|\sin\theta|$ and $|\cos\theta|$ equal $\sin\hat\theta$ and $\cos\hat\theta$; the signs are determined by ASTC.

Quadrant	Reference angle $\hat\theta$
I	$\theta$
II	$\pi - \theta$
III	$\theta - \pi$
IV	$2\pi - \theta$

Remarks

The unit circle is the gateway to everything deeper in trigonometry: the graphs of sine and cosine as periodic functions, the complex exponential $e^{i\theta} = \cos\theta + i\sin\theta$ (Euler's formula), Fourier series, and the geometry of rotations in any dimension. Whenever you encounter a trigonometric expression, ask yourself: where on the unit circle does this live, and what does the geometry tell me? That habit will make most competition trig problems feel almost effortless.

Practice Problems