The Unit Circle

Extend trigonometry beyond triangles using a circle of radius 1

If the unit circle seems scary to you, then know that you are not alone. When most people first hear “unit circle,” they picture some intimidating diagram covered with strange symbols, fractions with square roots, and angles written in ways they have never seen before. But here is the thing: the unit circle is just a simple circle. And you need to realize something: you already know how to do this. You may not believe it, but, really, you do.

Have you ever watched the second hand on a clock sweep around in a circle? Have you ever noticed a point on a spinning wheel or a Ferris wheel going around and around? That is the unit circle in action. The only “trick” is that mathematicians chose a circle with a very convenient size (radius equal to 1) and put it in a very convenient place (centered at the origin of a coordinate plane). Once you see why they made those choices, everything else falls into place naturally.

Core Concepts

What is the Unit Circle?

A unit circle is simply a circle with a radius of 1, centered at the origin of a coordinate plane. That is it. Nothing more complicated than that.

The word “unit” here means “one” - it is a circle where every point is exactly 1 unit away from the center. If you have worked with the coordinate plane before, you know that the equation $x^2 + y^2 = r^2$ describes a circle centered at the origin with radius $r$. When $r = 1$, this becomes:

$$x^2 + y^2 = 1$$

That is the equation of the unit circle. Any point $(x, y)$ that satisfies this equation lies on the unit circle.

Why Use a Circle with Radius 1?

You might be wondering: why this particular circle? Why not radius 2, or 5, or 10? The answer is that radius 1 makes calculations incredibly clean. When you work with a radius of 1, you do not have to multiply or divide by anything to convert between the circle’s measurements and the actual values you want. The coordinates of a point on the circle are directly the values you need.

Think of it like this: using a unit circle is like measuring with a ruler where every mark represents exactly 1 unit. No conversions, no scaling, just direct reading.

Coordinates on the Unit Circle: (cos θ, sin θ)

Here is where things get powerful. For any angle $θ$ measured from the positive x-axis, the point where the terminal side of that angle crosses the unit circle has coordinates:

$$(x, y) = (\cos θ, \sin θ)$$

Read that again, because it is the key to everything: the x-coordinate is the cosine of the angle, and the y-coordinate is the sine of the angle.

This means:

  • $\cos θ$ = the horizontal distance from the origin to the point
  • $\sin θ$ = the vertical distance from the origin to the point

For any angle, you can find the sine and cosine just by locating that point on the unit circle and reading off its coordinates. No triangles required (although triangles are secretly hiding in there, as we will see).

Why Does This Work? The Connection to Right Triangles

If you have studied right triangle trigonometry, you learned that for an angle $θ$ in a right triangle:

$$\sin θ = \frac{\text{opposite}}{\text{hypotenuse}} \quad \text{and} \quad \cos θ = \frac{\text{adjacent}}{\text{hypotenuse}}$$

Now imagine placing a right triangle inside the unit circle. The hypotenuse goes from the origin to a point on the circle. Since the radius of the unit circle is 1, the hypotenuse has length 1. This means:

$$\sin θ = \frac{\text{opposite}}{1} = \text{opposite} = y$$ $$\cos θ = \frac{\text{adjacent}}{1} = \text{adjacent} = x$$

When the hypotenuse equals 1, the opposite side equals sine and the adjacent side equals cosine directly. The unit circle does not replace triangles; it just makes the triangle calculations automatic by choosing a hypotenuse of 1.

Standard Position for Angles

Before we go further, let us establish how we measure angles on the unit circle. An angle is in standard position when:

  • Its vertex is at the origin
  • Its initial side (starting side) lies along the positive x-axis
  • Its terminal side (ending side) is where we measure to

We measure angles counterclockwise from the positive x-axis. This is considered the positive direction. If you measure clockwise, that is the negative direction.

So an angle of $90°$ points straight up. An angle of $180°$ points straight left. An angle of $270°$ points straight down. And an angle of $360°$ brings you all the way back to where you started.

The Four Quadrants

The coordinate plane is divided into four regions called quadrants. Since the unit circle sits at the center of this plane, understanding the quadrants helps you quickly determine the signs of trigonometric functions.

Quadrant I (upper right): Both x and y are positive. This covers angles from $0°$ to $90°$.

Quadrant II (upper left): x is negative, y is positive. This covers angles from $90°$ to $180°$.

Quadrant III (lower left): Both x and y are negative. This covers angles from $180°$ to $270°$.

Quadrant IV (lower right): x is positive, y is negative. This covers angles from $270°$ to $360°$.

Since $\cos θ = x$ and $\sin θ = y$, the signs of sine and cosine depend entirely on which quadrant your angle lands in.

Signs of Trigonometric Functions by Quadrant

Here is a pattern worth memorizing. In each quadrant, certain trig functions are positive:

Quadrant Positive Functions Memory Aid
I All (sin, cos, tan) All
II Sine only Students
III Tangent only Take
IV Cosine only Calculus

The mnemonic “All Students Take Calculus” (or ASTC) helps you remember which functions are positive in each quadrant, starting from Quadrant I and going counterclockwise.

Why does this work?

  • Quadrant I: x > 0, y > 0, so cos and sin are both positive, and tan = sin/cos is positive
  • Quadrant II: x < 0, y > 0, so cos is negative, sin is positive, and tan is negative
  • Quadrant III: x < 0, y < 0, so cos and sin are both negative, but tan = negative/negative is positive
  • Quadrant IV: x > 0, y < 0, so cos is positive, sin is negative, and tan is negative

Reference Angles

A reference angle is the acute angle (between $0°$ and $90°$) formed between the terminal side of an angle and the x-axis. Reference angles let you find trig values for any angle by connecting back to the familiar first-quadrant values.

Here is how to find a reference angle:

  • Quadrant I: Reference angle = $θ$
  • Quadrant II: Reference angle = $180° - θ$
  • Quadrant III: Reference angle = $θ - 180°$
  • Quadrant IV: Reference angle = $360° - θ$

For example, the reference angle for $150°$ is $180° - 150° = 30°$. This means $\sin 150°$ has the same absolute value as $\sin 30°$. You just need to check the sign based on the quadrant.

The Key Values to Memorize

You do not need to memorize the entire unit circle. You really only need to know the trig values for a few key angles in the first quadrant. Everything else can be derived from these using reference angles and quadrant signs.

Here are the essential first-quadrant values:

Angle $\sin θ$ $\cos θ$ $\tan θ$
$0°$ ($0$ rad) $0$ $1$ $0$
$30°$ ($\frac{π}{6}$ rad) $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
$45°$ ($\frac{π}{4}$ rad) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ $1$
$60°$ ($\frac{π}{3}$ rad) $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$
$90°$ ($\frac{π}{2}$ rad) $1$ $0$ undefined

Notice a pattern? As the angle increases from $0°$ to $90°$:

  • Sine goes from $0$ to $1$ (moving up through $\frac{1}{2}$, $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$)
  • Cosine goes from $1$ to $0$ (moving down through $\frac{\sqrt{3}}{2}$, $\frac{\sqrt{2}}{2}$, $\frac{1}{2}$)

The sine column and cosine column contain the same values, just in opposite order. This is because sine and cosine are related by $\sin θ = \cos(90° - θ)$.

A Memory Trick for the First Quadrant

Here is a trick for remembering the sine values of $0°$, $30°$, $45°$, $60°$, and $90°$:

Write the pattern: $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$

Simplifying: $0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1$

That is exactly the sine values for $0°$, $30°$, $45°$, $60°$, and $90°$! For cosine, just reverse the order.

Angles Beyond 360° and Negative Angles

The unit circle repeats every $360°$ (or $2π$ radians). This means:

$$\sin(θ + 360°) = \sin θ$$ $$\cos(θ + 360°) = \cos θ$$

So if you have an angle like $405°$, you can subtract $360°$ to get $45°$, and work with that simpler angle. Similarly, $750° = 750° - 2(360°) = 30°$.

For negative angles, you measure clockwise instead of counterclockwise. An angle of $-45°$ is the same position as $315°$ (since $360° - 45° = 315°$). To convert a negative angle to a positive one, keep adding $360°$ until you get a positive result.

Finding Trig Values Without a Calculator

With the unit circle, you can find exact values for many angles without any calculator. Here is the strategy:

  1. Identify the quadrant your angle is in
  2. Find the reference angle (the acute angle to the x-axis)
  3. Look up or recall the trig values for that reference angle
  4. Apply the correct sign based on the quadrant (ASTC)

This approach works for any angle that is a multiple of $30°$ or $45°$, which covers most of what you will encounter in trigonometry courses.

Notation and Terminology

Term Meaning Example
Unit circle Circle with radius 1 centered at the origin $x^2 + y^2 = 1$
Quadrant I Upper right (x > 0, y > 0) $0° < θ < 90°$
Quadrant II Upper left (x < 0, y > 0) $90° < θ < 180°$
Quadrant III Lower left (x < 0, y < 0) $180° < θ < 270°$
Quadrant IV Lower right (x > 0, y < 0) $270° < θ < 360°$
Standard position Angle with vertex at origin, initial side on positive x-axis All unit circle angles
Terminal side The ray that rotates to form an angle The side you measure to
Reference angle Acute angle to the x-axis Reference angle for $150°$ is $30°$
ASTC Memory aid: All Students Take Calculus Tells which trig functions are positive
Coterminal angles Angles that share the same terminal side $45°$ and $405°$ are coterminal

Examples

Example 1: Finding Coordinates on the Unit Circle

Find the coordinates of the point on the unit circle at $θ = 90°$.

Solution:

At $θ = 90°$, the terminal side points straight up along the positive y-axis.

On the unit circle, this point is 1 unit directly above the origin.

Therefore, the coordinates are: $$(x, y) = (0, 1)$$

This tells us:

  • $\cos 90° = 0$ (the x-coordinate)
  • $\sin 90° = 1$ (the y-coordinate)

This makes intuitive sense: at $90°$, you have moved all the way up (maximum y) but have no horizontal displacement (x = 0).

Example 2: Identifying the Quadrant

In which quadrant is $θ = 200°$?

Solution:

Recall the quadrant boundaries:

  • Quadrant I: $0°$ to $90°$
  • Quadrant II: $90°$ to $180°$
  • Quadrant III: $180°$ to $270°$
  • Quadrant IV: $270°$ to $360°$

Since $180° < 200° < 270°$, the angle $θ = 200°$ is in Quadrant III.

In Quadrant III, both x and y are negative, which means:

  • $\cos 200° < 0$
  • $\sin 200° < 0$
  • $\tan 200° > 0$ (negative divided by negative)

This matches our ASTC rule: only tangent is positive in Quadrant III (“T” for Take).

Example 3: Using Reference Angles

Find $\sin 150°$ and $\cos 150°$ using reference angles.

Solution:

Step 1: Identify the quadrant.

Since $90° < 150° < 180°$, the angle is in Quadrant II.

Step 2: Find the reference angle.

For Quadrant II, reference angle = $180° - θ = 180° - 150° = 30°$.

Step 3: Find the trig values for the reference angle.

From our key values:

  • $\sin 30° = \frac{1}{2}$
  • $\cos 30° = \frac{\sqrt{3}}{2}$

Step 4: Apply the correct signs.

In Quadrant II (upper left), x is negative and y is positive.

  • Sine is positive (y-coordinate): $\sin 150° = +\frac{1}{2}$
  • Cosine is negative (x-coordinate): $\cos 150° = -\frac{\sqrt{3}}{2}$

Final answer: $$\sin 150° = \frac{1}{2} \qquad \cos 150° = -\frac{\sqrt{3}}{2}$$

The coordinates of the point at $150°$ on the unit circle are $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

Example 4: Determining the Sign of a Trig Function

Determine the sign of $\tan 315°$.

Solution:

Step 1: Identify the quadrant.

Since $270° < 315° < 360°$, the angle is in Quadrant IV.

Step 2: Recall the ASTC rule.

In Quadrant IV, only Cosine is positive. This means:

  • $\cos 315° > 0$
  • $\sin 315° < 0$

Step 3: Find the sign of tangent.

Since $\tan θ = \frac{\sin θ}{\cos θ}$:

$$\tan 315° = \frac{\sin 315°}{\cos 315°} = \frac{\text{negative}}{\text{positive}} = \text{negative}$$

Final answer: $\tan 315° < 0$ (negative)

If you want the actual value: The reference angle is $360° - 315° = 45°$, and $\tan 45° = 1$, so $\tan 315° = -1$.

Example 5: Finding All Six Trig Functions

Find all six trigonometric functions for $θ = \frac{5π}{6}$.

Solution:

First, let us convert to degrees to help visualize: $\frac{5π}{6} \times \frac{180°}{π} = 150°$.

Step 1: Identify the quadrant.

Since $\frac{π}{2} < \frac{5π}{6} < π$ (or $90° < 150° < 180°$), this angle is in Quadrant II.

Step 2: Find the reference angle.

Reference angle = $π - \frac{5π}{6} = \frac{6π}{6} - \frac{5π}{6} = \frac{π}{6}$ (which is $30°$).

Step 3: Write the trig values for the reference angle $\frac{π}{6}$.

From our key values:

  • $\sin \frac{π}{6} = \frac{1}{2}$
  • $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$

Step 4: Apply Quadrant II signs.

In Quadrant II, sine is positive and cosine is negative.

$$\sin \frac{5π}{6} = \frac{1}{2}$$ $$\cos \frac{5π}{6} = -\frac{\sqrt{3}}{2}$$

Step 5: Find the remaining four trig functions.

$$\tan \frac{5π}{6} = \frac{\sin \frac{5π}{6}}{\cos \frac{5π}{6}} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{2} \times \frac{-2}{\sqrt{3}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$

$$\csc \frac{5π}{6} = \frac{1}{\sin \frac{5π}{6}} = \frac{1}{\frac{1}{2}} = 2$$

$$\sec \frac{5π}{6} = \frac{1}{\cos \frac{5π}{6}} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$

$$\cot \frac{5π}{6} = \frac{1}{\tan \frac{5π}{6}} = \frac{1}{-\frac{\sqrt{3}}{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3}$$

Final answers:

Function Value
$\sin \frac{5π}{6}$ $\frac{1}{2}$
$\cos \frac{5π}{6}$ $-\frac{\sqrt{3}}{2}$
$\tan \frac{5π}{6}$ $-\frac{\sqrt{3}}{3}$
$\csc \frac{5π}{6}$ $2$
$\sec \frac{5π}{6}$ $-\frac{2\sqrt{3}}{3}$
$\cot \frac{5π}{6}$ $-\sqrt{3}$
Example 6: Finding an Angle Given Conditions

Find $θ$ if $\cos θ = -\frac{\sqrt{3}}{2}$ and $θ$ is in Quadrant III.

Solution:

Step 1: Find the reference angle.

We need an angle whose cosine has absolute value $\frac{\sqrt{3}}{2}$.

From our key values, $\cos 30° = \frac{\sqrt{3}}{2}$.

So the reference angle is $30°$ (or $\frac{π}{6}$).

Step 2: Find the actual angle in Quadrant III.

For an angle in Quadrant III with reference angle $30°$:

$$θ = 180° + 30° = 210°$$

In radians: $θ = π + \frac{π}{6} = \frac{6π}{6} + \frac{π}{6} = \frac{7π}{6}$

Step 3: Verify the answer.

Check that $\cos 210° = -\frac{\sqrt{3}}{2}$:

  • Quadrant III: cosine should be negative ✓
  • Reference angle $30°$: $|\cos 210°| = \cos 30° = \frac{\sqrt{3}}{2}$ ✓

Therefore $\cos 210° = -\frac{\sqrt{3}}{2}$ ✓

Final answer: $θ = 210°$ or $θ = \frac{7π}{6}$ radians

Key Properties and Rules

Fundamental Unit Circle Relationship

For any angle $θ$, if $(x, y)$ is the corresponding point on the unit circle: $$x = \cos θ \qquad y = \sin θ$$

The Pythagorean Identity

Since every point on the unit circle satisfies $x^2 + y^2 = 1$, and since $x = \cos θ$ and $y = \sin θ$:

$$\cos^2 θ + \sin^2 θ = 1$$

This is called the Pythagorean identity and is one of the most important equations in trigonometry.

Quadrant Sign Rules (ASTC)

Quadrant sin cos tan
I + + +
II +
III +
IV +

Reference Angle Formulas

For an angle $θ$ in standard position:

Quadrant Reference Angle
I $θ$
II $180° - θ$ or $π - θ$
III $θ - 180°$ or $θ - π$
IV $360° - θ$ or $2π - θ$

Coterminal Angles

Angles that differ by a multiple of $360°$ (or $2π$) are coterminal - they share the same terminal side and have the same trig values:

$$\sin(θ + 360°n) = \sin θ$$ $$\cos(θ + 360°n) = \cos θ$$

where $n$ is any integer.

Converting Negative Angles

To find a positive coterminal angle for a negative angle: $$θ_{\text{positive}} = θ + 360°$$

Keep adding $360°$ until the result is positive.

Coordinates of Key Points

Angle (degrees) Angle (radians) Point on Unit Circle
$0°$ $0$ $(1, 0)$
$30°$ $\frac{π}{6}$ $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$45°$ $\frac{π}{4}$ $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$
$60°$ $\frac{π}{3}$ $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$90°$ $\frac{π}{2}$ $(0, 1)$
$180°$ $π$ $(-1, 0)$
$270°$ $\frac{3π}{2}$ $(0, -1)$

Real-World Applications

Circular Motion: Position on a Wheel

Imagine a point on the edge of a bicycle wheel with radius 1 foot. As the wheel rotates, the position of that point can be described using the unit circle.

If the wheel starts with the point at the rightmost position (3 o’clock) and rotates counterclockwise, after rotating through an angle $θ$:

  • Horizontal position: $x = \cos θ$ feet from the center
  • Vertical position: $y = \sin θ$ feet from the center

If the wheel has a different radius $r$, you just multiply: the position becomes $(r\cos θ, r\sin θ)$. This is why engineers, physicists, and animators love the unit circle - it provides a clean template that scales to any circular motion.

Periodic Phenomena: Seasons and Tides

Many natural phenomena repeat in cycles, and these cycles are described beautifully by the sine and cosine functions that come from the unit circle.

The tides rise and fall roughly twice a day in a pattern that can be modeled with sine functions. The height of the sun above the horizon throughout the year follows a similar pattern. Even the temperature variations between summer and winter can be approximated using sinusoidal functions.

When scientists say a phenomenon is “periodic,” they often mean it can be described using sine and cosine - and those functions come straight from the unit circle.

Computer Graphics: Rotating Objects

When a video game needs to rotate a spaceship, or when animation software spins a logo, the underlying math uses the unit circle.

To rotate a point $(x, y)$ by an angle $θ$ around the origin, you use:

$$x’ = x\cos θ - y\sin θ$$ $$y’ = x\sin θ + y\cos θ$$

These formulas come directly from the unit circle. Every spinning object you see on a screen, every rotating 3D model, every clock animation - all powered by the unit circle.

Signal Processing: Understanding Waves

Radio waves, sound waves, and electrical signals are often described as combinations of sine and cosine waves. Engineers use the unit circle to understand how these signals behave, combine, and interact.

When you tune into a radio station, your radio is essentially picking out signals with specific frequencies - frequencies that are described mathematically using the same sine and cosine functions you learn from the unit circle.

Self-Test Problems

Problem 1: What are the coordinates of the point on the unit circle at $θ = 180°$?

Show Answer

At $180°$, the terminal side points directly left along the negative x-axis.

The point is 1 unit to the left of the origin: $(-1, 0)$

This tells us $\cos 180° = -1$ and $\sin 180° = 0$.

Problem 2: In which quadrant is $θ = 325°$?

Show Answer

Since $270° < 325° < 360°$, the angle is in Quadrant IV.

In Quadrant IV, cosine is positive and sine is negative.

Problem 3: Find $\sin 225°$ and $\cos 225°$ without a calculator.

Show Answer

Step 1: $225°$ is in Quadrant III (since $180° < 225° < 270°$).

Step 2: Reference angle = $225° - 180° = 45°$.

Step 3: From key values: $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$.

Step 4: In Quadrant III, both sine and cosine are negative.

Final answer: $$\sin 225° = -\frac{\sqrt{2}}{2}$$ $$\cos 225° = -\frac{\sqrt{2}}{2}$$

Problem 4: Find the six trig functions for $θ = \frac{4π}{3}$.

Show Answer

First, convert to degrees: $\frac{4π}{3} \times \frac{180°}{π} = 240°$.

Quadrant: III (since $180° < 240° < 270°$)

Reference angle: $240° - 180° = 60°$ (or $\frac{π}{3}$)

From key values for $60°$: $\sin 60° = \frac{\sqrt{3}}{2}$, $\cos 60° = \frac{1}{2}$

In Quadrant III, sine and cosine are both negative:

$$\sin \frac{4π}{3} = -\frac{\sqrt{3}}{2}$$ $$\cos \frac{4π}{3} = -\frac{1}{2}$$ $$\tan \frac{4π}{3} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$ $$\csc \frac{4π}{3} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$ $$\sec \frac{4π}{3} = -2$$ $$\cot \frac{4π}{3} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

Problem 5: Find $\cos 510°$ without a calculator.

Show Answer

First, find a coterminal angle between $0°$ and $360°$: $$510° - 360° = 150°$$

Now find $\cos 150°$:

  • Quadrant II (since $90° < 150° < 180°$)
  • Reference angle: $180° - 150° = 30°$
  • $\cos 30° = \frac{\sqrt{3}}{2}$
  • In Quadrant II, cosine is negative

Final answer: $\cos 510° = \cos 150° = -\frac{\sqrt{3}}{2}$

Problem 6: If $\sin θ = \frac{\sqrt{2}}{2}$ and $θ$ is in Quadrant II, find $θ$.

Show Answer

Step 1: Find the reference angle.

$\sin 45° = \frac{\sqrt{2}}{2}$, so the reference angle is $45°$.

Step 2: Find the angle in Quadrant II.

For Quadrant II: $θ = 180° - 45° = 135°$

Step 3: Verify.

  • $135°$ is in Quadrant II ✓
  • $\sin 135° = +\frac{\sqrt{2}}{2}$ (positive because Quadrant II) ✓

Final answer: $θ = 135°$ or $θ = \frac{3π}{4}$ radians

Problem 7: At what angle(s) between $0°$ and $360°$ does $\cos θ = 0$?

Show Answer

$\cos θ = 0$ means the x-coordinate on the unit circle is 0.

This happens when the terminal side points straight up or straight down:

  • Straight up: $θ = 90°$
  • Straight down: $θ = 270°$

Final answer: $θ = 90°$ and $θ = 270°$ (or $\frac{π}{2}$ and $\frac{3π}{2}$ radians)

Summary

  • The unit circle is a circle with radius 1 centered at the origin. Its equation is $x^2 + y^2 = 1$.

  • For any angle $θ$ in standard position, the point where the terminal side crosses the unit circle has coordinates $(\cos θ, \sin θ)$. The x-coordinate is the cosine, and the y-coordinate is the sine.

  • This works because of right triangle trigonometry: when the hypotenuse equals 1 (the radius of the unit circle), sine equals the opposite side (y) and cosine equals the adjacent side (x) directly.

  • The coordinate plane has four quadrants. Use ASTC (All Students Take Calculus) to remember which trig functions are positive in each quadrant:

    • Quadrant I: All positive
    • Quadrant II: Sine positive
    • Quadrant III: Tangent positive
    • Quadrant IV: Cosine positive
  • A reference angle is the acute angle to the x-axis. It lets you convert any angle to an equivalent first-quadrant problem, then apply the appropriate sign.

  • You only need to memorize trig values for $0°$, $30°$, $45°$, $60°$, and $90°$. Everything else can be found using reference angles and quadrant signs.

  • Coterminal angles differ by multiples of $360°$ (or $2π$) and have the same trig values. For negative angles, add $360°$ until positive.

  • The Pythagorean identity $\cos^2 θ + \sin^2 θ = 1$ comes directly from the unit circle equation.

  • The unit circle appears in real-world applications everywhere: circular motion (wheels, orbits), periodic phenomena (tides, seasons), computer graphics (rotations), and signal processing (waves).

The unit circle may look intimidating at first, but it is really just a beautifully organized way to extend trigonometry beyond triangles. Once you understand that every point on the circle gives you cosine and sine directly from its coordinates, and that reference angles let you relate any angle back to the familiar first-quadrant values, the entire circle becomes a tool you can use confidently. And unlike a calculator, it gives you exact values, not decimal approximations. That is the power of the unit circle.