Unit circle definition of trig functions — all 6 trig functions for any angle: Visual walkthrough

We will build the whole idea in numbered steps. Each step tells you WHAT we do, WHY we do it, and shows you WHAT IT LOOKS LIKE. Follow the red arrows.

Step 1 — A circle of radius one, and a walker on it

WHAT. Draw a flat plane. Mark two number lines crossing at a point called the origin (the spot $(0, 0)$ ). The horizontal line is the x-axis (how far right/left) and the vertical line is the y-axis (how far up/down). Now draw a circle centered at the origin whose radius — the distance from center to edge — is exactly $1$ . This is the unit circle ("unit" just means "one").

WHY. We want a machine that turns an angle into numbers. A circle is perfect: every point on it is the same distance from the center, so the only thing that changes as we move is direction. And direction is exactly what an angle measures. Choosing radius $1$ is a deliberate simplification — you'll see in Step 4 that it makes a messy fraction collapse to just "the coordinate."

PICTURE. Our walker always starts at the point $(1, 0)$ — the "3 o'clock" spot, one unit to the right. From there it will walk counterclockwise (the agreed positive direction, like a clock running backwards).

Step 2 — What an angle is here: how far you've walked

WHAT. Pick an amount of turn and call it $θ$ (the Greek letter "theta" — just a name for "the angle"). Rotate the walker counterclockwise by $θ$ , starting from $(1, 0)$ . It stops somewhere on the circle. Call that stopping point $P$ .

WHY. We need a single label for "how much we turned" so we can talk about it. $θ$ is that label. Notice we have not yet used any triangle — $θ$ is defined purely as rotation around the circle. This is the key upgrade over school trig: a triangle can only hold angles between $0°$ and $90°$ , but you can keep walking around a circle forever.

PICTURE. The angle $θ$ is the wedge swept out between the starting arm (pointing to $(1, 0)$ ) and the final arm (pointing to $P$ ). The little arc shows the sweep.

Step 3 — Name the landing point's coordinates

WHAT. The walker has stopped at $P$ . Drop a straight vertical line from $P$ down to the x-axis, and a straight horizontal line from $P$ across to the y-axis. Read off the two coordinates: call them $x$ and $y$ .

WHY. These two numbers completely pin down where you are. Instead of describing your position with the vague word "there," we now have two exact numbers. The whole plan of the unit circle is: turn the angle $θ$ into these two numbers.

PICTURE. The blue segment along the bottom is $x$ (horizontal reach); the orange segment going up is $y$ (vertical reach). Together with the radius they will form a triangle in Step 4.

Step 4 — The hidden right triangle, and why $cos θ = x$

WHAT. Look at three line segments: the radius from origin to $P$ , the horizontal segment of length $x$ , and the vertical segment of length $y$ . They form a right triangle (the corner where $x$ meets $y$ is a $90°$ square corner).

WHY. This connects the new circle picture to old school trig, Right-triangle trigonometry (SOH-CAH-TOA). In a right triangle, cosine is defined as $cos θ = \frac{adjacent}{hypotenuse} .$ Let's read every symbol on our triangle:

hypotenuse ::: the long slanted side = the radius = $1$ .
adjacent ::: the side next to the angle $θ$ that isn't the hypotenuse = the horizontal side = $x$ .
opposite ::: the side across from $θ$ = the vertical side = $y$ .

Now substitute: $cos θ = \frac{adjacent}{hypotenuse} = \frac{x}{1} = x .$ The denominator is $1$ — this is why we chose radius $1$ . The division does nothing, and cosine simply becomes the x-coordinate. Identically, $sin θ = \frac{opposite}{hypotenuse} = \frac{y}{1} = y .$

PICTURE. The triangle sits inside the circle. Hypotenuse in gray ( $= 1$ ), adjacent leg in blue ( $= x = cos θ$ ), opposite leg in orange ( $= y = sin θ$ ).

Step 5 — The other four functions are just ratios of these two

WHAT. Build the remaining four trig functions purely from $x$ and $y$ : $tan θ = \frac{y}{x}, cot θ = \frac{x}{y}, sec θ = \frac{1}{x}, csc θ = \frac{1}{y} .$

WHY. We don't invent four new ideas — that would be four things to memorize. Instead each is a ratio of the two we already have. Reading each symbol:

$tan θ = \frac{y}{x} = \frac{sin θ}{cos θ}$ — "opposite over adjacent," the steepness (rise over run) of the radius arm.
$cot θ = \frac{x}{y}$ — tangent flipped upside down.
$sec θ = \frac{1}{x} = \frac{1}{cos θ}$ — reciprocal of cosine.
$csc θ = \frac{1}{y} = \frac{1}{sin θ}$ — reciprocal of sine.

PICTURE. The green line shows $tan θ$ as the slope of the radius: how many units up ( $y$ ) per unit right ( $x$ ). A steeper arm means a bigger $tan$ .

Step 6 — The degenerate cases: when a coordinate hits zero

WHAT. Watch what happens when the walker lands exactly on an axis, so one coordinate becomes $0$ .

WHY. Dividing by zero is not allowed — it produces no number at all. So any function with a zero in the denominator is undefined there. This is not a flaw; it's the geometry telling you the ratio has no answer.

At $θ = 90°$ and $270°$ : the point is on the y-axis, so $x = cos θ = 0$ . Then $tan θ = \frac{y}{0}$ and $sec θ = \frac{1}{0}$ are undefined.
At $θ = 0°$ and $180°$ : the point is on the x-axis, so $y = sin θ = 0$ . Then $cot θ = \frac{x}{0}$ and $csc θ = \frac{1}{0}$ are undefined.

PICTURE. Two landing points shown: top ( $x = 0$ , breaks $tan, sec$ ) and right ( $y = 0$ , breaks $cot, csc$ ). The red "÷0" marks the broken ratio.

Step 7 — The Pythagorean identity is the circle equation in disguise

WHAT. Take the circle's own equation $x^{2} + y^{2} = 1$ and replace $x$ with $cos θ$ and $y$ with $sin θ$ : $cos^{2} θ + sin^{2} θ = 1.$

WHY. The point $P$ is on the circle, so it must satisfy the circle's equation. There is nothing to prove — the identity is the circle wearing different clothes. Reading it: $cos^{2} θ$ is the horizontal reach squared, $sin^{2} θ$ is the vertical reach squared, and their sum is the radius squared $= 1^{2} = 1$ . This is just the Pythagorean theorem on the Step 4 triangle. (See Pythagorean identities for the two spin-offs found by dividing through by $cos^{2} θ$ or $sin^{2} θ$ .)

PICTURE. The right triangle labelled with legs $cos θ$ , $sin θ$ and hypotenuse $1$ , with the squared-lengths adding up.

Step 8 — Every quadrant: how the signs flip

WHAT. As the walker rounds the circle it enters four regions called quadrants. In each, $x$ and $y$ carry different signs, so the trig functions do too.

WHY. School triangles only live in Quadrant I where everything is positive. The circle lets $x$ or $y$ go negative, and we must track that. Since $cos θ = x$ and $sin θ = y$ , the sign of each function is simply the sign of that coordinate. See Reference angles for finding the magnitude, then attach the sign from here.

Quadrant	walker is	$x = cos$	$y = sin$	positive functions
I	upper-right	$+$	$+$	All
II	upper-left	$-$	$+$	Sine (& csc)
III	lower-left	$-$	$-$	Tan (& cot)
IV	lower-right	$+$	$-$	Cos (& sec)

PICTURE. All four quadrants coloured, each labelled with the signs of $(x, y)$ and which functions stay positive ("All Students Take Calculus," counterclockwise from Q1).

Step 9 — Negative angles and full loops: symmetry and repetition

WHAT. Two special walks:

Negative angle $- θ$ = walk clockwise instead. You land at the mirror image across the x-axis: same $x$ , flipped $y$ . Hence $cos (- θ) = cos θ (even), sin (- θ) = - sin θ (odd) .$
Add $360°$ (one full loop, $2 π$ radians) and you return to the exact same point, so every function repeats: $sin (θ + 360°) = sin θ, and so on.$

WHY. "Even/odd" (Even and odd functions) and "periodic" (Periodicity and $2\pi$) sound abstract, but on the circle they are just reflection and coming back around. Reading the equations: the minus sign in $sin (- θ) = - sin θ$ is exactly the $y$ -coordinate flipping below the axis; the " $+ 360°$ does nothing" is the loop closing. These are why the Graphs of trig functions wave up and down forever with mirror symmetry.

PICTURE. Left: $+ θ$ and $- θ$ as mirror points sharing an $x$ . Right: $θ$ and $θ + 360°$ landing on the identical point.

The one-picture summary

Everything above compressed into a single diagram: the walker at angle $θ$ , its coordinates $x = cos θ$ (blue) and $y = sin θ$ (orange), the slope $tan θ$ (green), the hypotenuse $1$ closing the Pythagorean triangle, and the quadrant sign map around the rim.

Recall Feynman: the whole walkthrough in plain words

Picture a one-metre merry-go-round; you climb on at the 3 o'clock spot. Spin some amount $θ$ . Now ask two questions about where you sit: how far right am I? — that number is $cos θ$ . How far up am I? — that's $sin θ$ . Drop a plumb line and a level line and you've drawn a right triangle whose slanted side is the one-metre radius; because the radius is exactly one, "adjacent over hypotenuse" is just "right-ness over one," which is right-ness itself — that's why cosine literally equals the x-coordinate. The four other functions are only these two shuffled: their ratio ( $tan =$ up-over-right, the steepness), and their upside-downs. When you sit exactly on top or on the side, one of the numbers is zero, and dividing by it "breaks the calculator" — that's undefined. Cross to the left side and "how far right" turns negative, which is where the quadrant signs come from. Spin backwards and you mirror below the line (sine flips, cosine doesn't). Spin a whole loop and you're right back where you started — so the numbers repeat forever. And since you never left the circle, right-squared plus up-squared always equals one: that's the Pythagorean identity, free of charge.

Recall check

Which coordinate is

cos θ

The x-coordinate (horizontal reach) of the landing point.

Why does radius

1

make

cos θ = x

exactly?

Because

cos θ = adjacent / hyp = x /1 = x

; the hypotenuse is

1

so the division does nothing.

Why is

tan θ

undefined at

90°

The point is

(0, 1)

, so

x = cos 90° = 0

and

tan = y / x

divides by zero.

Where does

sin^{2} θ + cos^{2} θ = 1

come from?

It IS the circle equation

x^{2} + y^{2} = 1

with

x = cos θ

y = sin θ

Why is cosine even but sine odd?

A negative angle mirrors the point across the x-axis:

x

stays (cos even),

y

flips (sin odd).

In Quadrant III, which functions are positive?

Only

tan

and

cot

(both

x

and

y

are negative, so their ratio is positive).

Connections

Unit Circle Definition of Trig Functions — All 6 for Any Angle — the parent this page unpacks.
Right-triangle trigonometry (SOH-CAH-TOA) — Step 4's triangle is the bridge.
Reference angles — get the magnitude, then attach the Step 8 sign.
Pythagorean identities — Step 7 and its two divisions.
Radian measure — angle as arc length on the radius- $1$ circle.
Graphs of trig functions — plotting the $x$ and $y$ coordinates versus $θ$ .
Even and odd functions — Step 9's reflection symmetry.
Periodicity and $2\pi$ — Step 9's full-loop repetition.