The unit circle is the map that connects angles to coordinates, and coordinates to the trigonometric functions. Once you understand it, you can find the sine, cosine, and tangent of any angle, positive, negative, or larger than $360^\circ$, without a calculator.
Today's hook, Imagine a circle with radius 1 centred at the origin. If you walk around the circumference to an angle of 45° (π/4 radians), what are your x- and y-coordinates? How do these coordinates relate to sin 45° and cos 45°?
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
Imagine a circle with radius 1 centred at the origin. If you walk around the circumference to an angle of $45^\circ$ ($\frac{\pi}{4}$ radians), without using a calculator what are your $x$- and $y$-coordinates? How do these coordinates relate to $\sin 45^\circ$ and $\cos 45^\circ$?
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The unit circle identity
Work through the core explanation before applying it.
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The unit circle identity
+5 XP to read
There is only one identity at the heart of this lesson. Every trig value for every angle flows from it. Lock this into memory and the rest of trigonometry becomes navigable.
If you rotate by angle $\theta$ from $(1, 0)$ on a circle of radius 1, you land at the point $(\cos \theta, \sin \theta)$. The $x$-coordinate is cosine. The $y$-coordinate is sine. The gradient is tangent.
$P(\theta) = (\cos \theta, \sin \theta)$
Coordinates = (cos, sin)
$x = \cos \theta$ and $y = \sin \theta$. Cosine comes first in the alphabet, just like $x$ comes before $y$.
All positive in QI, Sin in QII, Tan in QIII, Cos in QIV. Tells you which ratios are positive where.
A point at angle $\theta$ on the unit circle has coordinates $(\cos\theta, \sin\theta)$; $x = \cos\theta$ (remember: C comes before S, just as x comes before y)
Pause, copy the unit circle coordinate identity: point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$, $x$ = cosine (C before S, just as x before y) into your book.
Did you get this? True or false: for a point on the unit circle at angle $\theta$, the $y$-coordinate equals $\cos\theta$.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The definition of the unit circle
How sine, cosine, and tangent are defined from the unit circle
The ASTC rule for quadrant signs
Key points on the unit circle at multiples of $\frac{\pi}{2}$
Understand
Concepts
Why the unit circle extends trig ratios to any angle
How reference angles simplify calculations in all quadrants
The connection between angles and coordinates
Can do
Skills
Find $\sin \theta$, $\cos \theta$, and $\tan \theta$ from the unit circle
Determine the quadrant of any angle in radians
Use reference angles to find exact values
Apply the Pythagorean identity to find missing trig ratios
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Key terms
We just saw that every point on the unit circle encodes $(\cos\theta, \sin\theta)$. That raises a question: but which angles give which signs, how do we know if cosine or sine is positive or negative in each quadrant? This card answers it → the ASTC rule and the reference angle concept, which let you evaluate any angle from just the acute cases.
Unit circleA circle with radius 1 centred at the origin $(0, 0)$.
QuadrantOne of four regions of the coordinate plane divided by the axes.
Reference angleThe acute angle between the terminal side and the $x$-axis. Always positive.
Terminal sideThe ray that rotates from the positive $x$-axis to form angle $\theta$.
ASTCMemory aid: All (QI), Sin (QII), Tan (QIII), Cos (QIV) are positive.
Pythagorean identity$\cos^2 \theta + \sin^2 \theta = 1$ for every angle $\theta$.
Unit circle: radius = 1, centred at origin, equation $x^2 + y^2 = 1$; Reference angle: always acute (between 0 and $\frac{\pi}{2}$), measured from the $x$-axis
Pause, copy the unit circle equation $x^2 + y^2 = 1$, the reference angle definition (always acute, from the $x$-axis), and the ASTC memory aid into your book.
Quick check: In which quadrant are both sine and cosine negative?
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The unit circle, what's actually going on
Work through the core explanation before applying it.
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The unit circle, what's actually going on
core concept
We just saw the ASTC rule for sign and the reference angle definition. That raises a question: how do reference angles combine with ASTC to let us evaluate any angle, including obtuse and reflex angles, using only the acute exact values? This card answers it → the four reference angle formulas (one per quadrant) applied to the unit circle.
The unit circle is a circle with radius $1$ centred at the origin $(0, 0)$. It is the foundation of trigonometry because it connects angles to coordinates in a simple, universal way.
If you start at the point $(1, 0)$ and rotate anticlockwise by an angle $\theta$, the point where you land is:
$$P(\theta) = (\cos \theta, \sin \theta)$$
This means the $x$-coordinate is $\cos \theta$, the $y$-coordinate is $\sin \theta$, and the gradient of the line from the origin is $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
The unit circle, radius 1, centred at the origin, showing key angles and the ASTC quadrants.
Reference angles
The reference angle $\alpha$ is the acute angle that the terminal side makes with the $x$-axis. It is always positive and between $0$ and $\frac{\pi}{2}$.
Q I
$\alpha = \theta$
Q II
$\alpha = \pi - \theta$
Q III
$\alpha = \theta - \pi$
Q IV
$\alpha = 2\pi - \theta$
Once you know the reference angle, find the exact trig value using special triangles, then apply the correct sign from ASTC.
Why GPS satellites use the unit circle. GPS satellites orbit the Earth in nearly circular paths. To calculate your position on the ground, the GPS receiver solves triangles using the sine and cosine of orbital angles, angles that can be any size, positive or negative, depending on the satellite's position relative to you. The unit circle extends these trig functions beyond acute angles, making global positioning possible.
Unit circle point: $P(\theta) = (\cos\theta, \sin\theta)$, x is cos, y is sin; Reference angle formulas: QI: $\alpha=\theta$, QII: $\alpha=\pi-\theta$, QIII: $\alpha=\theta-\pi$, QIV: $\alpha=2\pi-\theta$
Pause, copy the unit circle coordinate identity $P(\theta) = (\cos\theta, \sin\theta)$ and all four reference angle formulas (QI: $\alpha=\theta$; QII: $\alpha=\pi-\theta$; QIII: $\alpha=\theta-\pi$; QIV: $\alpha=2\pi-\theta$) into your book.
Fill the blanks: drag each token into the matching blank.
π − θθ − π2π − θθ
In QI the reference angle is ___.
In QII it is ___.
In QIII it is ___.
In QIV it is ___.
Predict then reveal+8 XP
1 · Predict
2 · Reveal
3 · Compare
How many angles $\theta$ in $[0, 2\pi)$ satisfy $\sin\theta = -0.6$, and which quadrants are they in?
50%
Actual answer
Two solutions, in Quadrant III and Quadrant IV.
Sine is negative below the $x$-axis, which is QIII and QIV. Both share the reference angle $\alpha = \arcsin(0.6) \approx 0.64$, giving $\theta \approx \pi + 0.64 = 3.79$ and $\theta \approx 2\pi - 0.64 = 5.64$. Their cosines have opposite signs, $-0.8$ and $+0.8$, which is why the quadrant has to decide the sign and the square root cannot. Find both in the explorer below.
How close was your prediction?
Good. Every sine value between $-1$ and $1$ has two solutions in one turn, never one.
A horizontal line at height $-0.6$ cuts the circle twice, so there are always two solutions in one full turn.
Interactive · The unit circle
Try this: drag the point, or focus the circle and use the arrow keys. Watch $\cos\theta$ and $\sin\theta$ move as the $x$- and $y$-coordinates. Then use find both solutions to hunt down every angle with a given sine, and the sign probe to let the quadrant choose the sign for you.
Use the explorer. In the hunt panel, find both solutions of $\sin\theta = -0.6$ in $[0, 2\pi)$. Read the two angles off the caption, in radians to two decimal places: the Quadrant III solution is and the Quadrant IV solution is .
Use the explorer. Park the point at the Quadrant IV solution above and read $\cos\theta$. Which sign is right, and why?
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · COORDINATES ON THE UNIT CIRCLE
Find the exact coordinates of the point on the unit circle corresponding to $\theta = \frac{2\pi}{3}$.
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Quadrant II ($\frac{\pi}{2} < \frac{2\pi}{3} < \pi$)
In QIV, cosine is positive. $\cos \theta = \frac{4}{5}$
Apply ASTC to choose the sign.
PROBLEM 3 · NEGATIVE ANGLES
Find the exact value of $\cos\left(-\frac{\pi}{4}\right)$.
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$-\frac{\pi}{4}$ means clockwise rotation from $(1, 0)$
Understand negative angles.
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Lands in Quadrant IV with reference angle $\frac{\pi}{4}$
Find the equivalent position.
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$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. In QIV, cosine is positive.
Apply value and sign.
Follow-up: Using the same approach, what is the exact value of $\sin\!\left(-\frac{\pi}{6}\right)$?
Common errors · the 3 traps that cost marks
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Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Swapping sine and cosine
Students often write $\sin \theta = x$ and $\cos \theta = y$. Remember: $x$ comes before $y$ in the alphabet, and $\cos$ comes before $\sin$ in the coordinate pair $(\cos \theta, \sin \theta)$.
Trap 02
Getting the reference angle formula wrong
In Quadrant II, the reference angle is $\pi - \theta$, not $\theta - \pi$. Some students subtract the wrong way and get negative reference angles. Reference angles are always positive and acute.
Trap 03
Forgetting ASTC when finding missing ratios
When using $\cos^2 \theta + \sin^2 \theta = 1$ to find a missing ratio, taking the square root gives two possible answers ($\pm$). You must use the quadrant information to choose the correct sign.
Odd one out: Three of these statements are correct. Which one is WRONG?
Quick-fire practice · 5 reps
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
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Find the coordinates on the unit circle for $\theta = \pi$.
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Find the coordinates on the unit circle for $\theta = \frac{5\pi}{4}$.
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Find the coordinates on the unit circle for $\theta = -\frac{\pi}{2}$.
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Find the coordinates on the unit circle for $\theta = \frac{5\pi}{6}$.
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State the exact value of $\tan \frac{4\pi}{3}$.
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Revisit your thinking
Earlier you were asked: What are your coordinates at $45^\circ$ on the unit circle, and how do they relate to $\sin 45^\circ$ and $\cos 45^\circ$?
At $\theta = \frac{\pi}{4}$ ($45^\circ$), the point on the unit circle is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. By the unit circle definition, the $x$-coordinate is $\cos \theta$ and the $y$-coordinate is $\sin \theta$. Therefore $\cos 45^\circ = \frac{\sqrt{2}}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$. This is the only acute angle where sine and cosine are exactly equal.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 4
Q1. (a) State the exact coordinates of the point on the unit circle corresponding to $\theta = \frac{3\pi}{4}$. (b) Hence, write down the exact values of $\sin \frac{3\pi}{4}$ and $\cos \frac{3\pi}{4}$. 3 MARKS
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ApplyBand 4
Q2. If $\cos \theta = -\frac{5}{13}$ and $\theta$ is in Quadrant II, find the exact value of $\sin \theta$. Show your working. 2 MARKS
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AnalyseBand 5
Q3. A student claims that $\sin \theta$ is always positive when $\theta$ is between $0$ and $\pi$, and always negative when $\theta$ is between $\pi$ and $2\pi$. Evaluate this claim, using specific examples from the unit circle to support your answer. 3 MARKS
Q2 (2 marks): $\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}$ [1]. In QII, sine is positive, so $\sin \theta = \frac{12}{13}$ [1].
Q3 (3 marks): The claim is partially correct but imprecise [0.5]. Between $0$ and $\pi$, sine is positive in QI and QII, so the first part is correct [0.5]. Between $\pi$ and $2\pi$, sine is negative in QIII and QIV, so the second part is also correct [0.5]. But the claim ignores that $\sin \pi = 0$ and $\sin 2\pi = 0$, which are neither positive nor negative [1]. The claim is broadly true for open intervals $(0, \pi)$ and $(\pi, 2\pi)$, but false at the endpoints [0.5].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
