Everything you know about sine and cosine so far lives inside a right-angled triangle, which means it only applies to acute angles. The unit circle rebuilds both definitions so they work for every angle there is, and it agrees with the old definitions wherever both apply.
Today's hook: What is $\sin 120°$? There is no right-angled triangle with a $120°$ angle in it, so the definition you have does not apply. Yet the number exists, your calculator returns it, and it matters. Getting a definition that reaches it is the job of this lesson.
0/5QUESTS
Think First
warm-up
Draw a circle of radius $1$ centred at the origin, and mark a point $P$ on it in the upper right. Drop a perpendicular from $P$ to the horizontal axis to make a right-angled triangle. What is the length of its hypotenuse, and what does that do to the fractions in SOH-CAH-TOA?
Record your answer in your workbook.
1
The Big Idea
+5 XP to read
On a circle of radius $1$ centred at the origin, let $P$ be the point reached by turning through an angle $\theta$ from the positive horizontal axis. Then $\cos\theta$ is the first coordinate of $P$ and $\sin\theta$ is the second.
$$P = (\cos\theta, \ \sin\theta)$$
The definition costs nothing where the old one worked and keeps working where it did not. For an acute $\theta$ the triangle in the diagram has hypotenuse $1$, so the fractions in SOH-CAH-TOA have denominator $1$ and reduce to the coordinates themselves.
$P = (\cos\theta, \sin\theta)$
Across then up
Cosine is the first coordinate, sine the second. They are in alphabetical order.
Angles turn anticlockwise
Measured from the positive horizontal axis, turning anticlockwise. That is the convention.
Never outside $-1$ to $1$
A coordinate of a point on a circle of radius $1$ cannot exceed $1$ in size.
2
What You'll Master
objectives
Know
That $\cos\theta$ and $\sin\theta$ are the coordinates of a point on the unit circle
That angles are measured anticlockwise from the positive horizontal axis
The values of sine and cosine at $0°$, $90°$, $180°$, $270°$ and $360°$
Understand
Why the new definition agrees with SOH-CAH-TOA for acute angles
Why sine and cosine can never lie outside $-1$ to $1$
Can Do
Locate the point on the unit circle for a given angle
Read sine and cosine values off the unit circle at the quadrant boundaries
Use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$
3
Words You Need
vocabulary
Unit circleThe circle of radius $1$ centred at the origin.
Standard positionAn angle measured anticlockwise from the positive horizontal axis, with its vertex at the origin.
QuadrantOne of the four regions the axes divide the plane into, numbered anticlockwise from the upper right.
Pythagorean identity$\sin^2\theta + \cos^2\theta = 1$, true for every angle.
FunctionSine and cosine assign exactly one output to each angle, so each is a function of the angle.
4
Why a New Definition Is Needed
+5 XP to read
The definitions you have are ratios of sides in a right-angled triangle:
They work, and they will keep working. But they carry a hidden restriction: the angles of a right-angled triangle other than the right angle are always acute, so these definitions say nothing at all about $\theta = 120°$, or $\theta = 250°$, or $\theta = -40°$.
That restriction has to go, for two reasons. Some real quantities genuinely involve obtuse angles: a triangle with an obtuse angle still has an area, and the sine rule still applies to it. And beyond Year 10, sine and cosine are used to describe anything that repeats — tides, sound, alternating current, seasons — where the angle runs on indefinitely and is not the angle of any triangle at all.
A good replacement must agree with the old definition wherever the old one applied, or every previous result would be at risk. The unit circle definition does exactly that, and the next card shows why.
5
The Definition
+5 XP to read
Draw the circle of radius $1$ centred at the origin: the unit circle.
Start at the point $(1, 0)$ and turn anticlockwise through an angle $\theta$, staying on the circle. Call the point you reach $P$. Then, by definition:
$$\cos\theta = \text{the first coordinate of } P, \qquad \sin\theta = \text{the second coordinate of } P$$
so that $P = (\cos\theta, \ \sin\theta)$.
Two conventions are built in and are worth stating explicitly. Angles are measured from the positive horizontal axis, and anticlockwise is positive. Turning clockwise gives a negative angle, which Lesson 3 takes up.
Nothing in this definition mentions a triangle, so nothing restricts $\theta$. Turning through $120°$ lands you somewhere in the upper left; turning through $250°$ lands you in the lower left; turning through $400°$ takes you a full lap and $40°$ more. Every angle reaches a point, so every angle has a sine and a cosine.
Order matters
Cosine first, sine second. A useful memory: the two names are in alphabetical order, and so are the two coordinates. Swapping them is the single most common error in this whole focus area.
6
It Agrees With What You Knew
+5 XP to read
Take an acute angle $\theta$ and drop a perpendicular from $P$ to the horizontal axis, as in the diagram. This makes a right-angled triangle with:
hypotenuse the radius, of length $1$; the side adjacent to $\theta$ of length equal to the first coordinate of $P$; the side opposite $\theta$ of length equal to the second coordinate.
which is exactly the new definition. The radius being $1$ is what makes the denominators disappear, and that is the whole reason the circle is chosen to have radius $1$.
So the new definition is a genuine extension: it does not replace what you knew, it continues it. Every result proved with SOH-CAH-TOA remains valid, and the new definition reaches angles the old one could not.
7
What the Circle Tells You Immediately
+5 XP to read
Three facts fall out of the definition with no work at all.
The values at the quadrant boundaries. Read the coordinates of the four points where the circle meets the axes:
$\theta = 0°$ gives $P = (1,0)$, so $\cos 0° = 1$ and $\sin 0° = 0$; $\theta = 90°$ gives $P = (0,1)$, so $\cos 90° = 0$ and $\sin 90° = 1$; $\theta = 180°$ gives $P = (-1,0)$; $\theta = 270°$ gives $P = (0,-1)$; $\theta = 360°$ gives $P = (1,0)$ again, back where it started.
The range. A point on a circle of radius $1$ has both coordinates between $-1$ and $1$, so
for every angle. An answer outside that range is wrong, which makes this a fast check on any calculation.
The Pythagorean identity. The point $(\cos\theta, \sin\theta)$ is at distance $1$ from the origin, so by Pythagoras:
$$\cos^2\theta + \sin^2\theta = 1$$
This holds for every angle, not just acute ones, because the point is on the circle for every angle. It is the most-used identity in senior trigonometry, and here it is simply the equation of the circle.
8
Common Pitfalls
+5 XP to read
Swapping the two coordinates, taking sine as the first and cosine as the second.
Fix: cosine is the first coordinate, in alphabetical order with the axes. Check against $\theta = 0°$: the point is $(1,0)$, and $\cos 0° = 1$, so cosine must be the first.
Measuring the angle from the vertical axis, or clockwise.
Fix: from the positive horizontal axis, turning anticlockwise. A clockwise turn is a negative angle, which means something different.
Giving a sine or cosine value greater than $1$.
Fix: impossible. Both are coordinates of a point on a circle of radius $1$, so both lie between $-1$ and $1$. Check every answer against this.
Assuming the unit circle replaces SOH-CAH-TOA and that right-triangle work is now wrong.
Fix: it extends it. For acute angles the two definitions give identical answers, which is exactly why the extension is allowed.
Watch Me Solve It · 3 examples
Watch Me Solve It · Reading values off the circle
+15 XP per step
Q1
PROBLEM
Use the unit circle to state the exact values of (a) $\sin 180°$, (b) $\cos 270°$, (c) $\sin 90°$, (d) $\cos 360°$.
1
(a) Locate the point for 180 degrees
$P = (-1, 0)$
Half a turn anticlockwise from $(1,0)$ lands on the negative horizontal axis. The second coordinate is $0$, so $\sin 180° = 0$.
2
(b) Locate the point for 270 degrees
$P = (0, -1)$
Three quarters of a turn lands at the bottom of the circle. The first coordinate is $0$, so $\cos 270° = 0$.
3
(c) Locate the point for 90 degrees
$P = (0, 1)$
A quarter turn lands at the top. The second coordinate is $1$, so $\sin 90° = 1$.
4
(d) Recognise a full turn
$P = (1, 0)$
A full turn returns to the starting point, so $\cos 360° = 1$, the same as $\cos 0°$. This is the first sign that the functions repeat, which Lesson 4 develops.
Nice work, XP earned
Answer(a) $0$; (b) $0$; (c) $1$; (d) $1$
Watch Me Solve It · Using the identity
+15 XP per step
Q2
PROBLEM
An angle $\theta$ in the first quadrant has $\cos\theta = 0.6$. Find $\sin\theta$ exactly, and state the coordinates of the corresponding point on the unit circle.
1
Write the identity
$\cos^2\theta + \sin^2\theta = 1$
True for every angle, since the point lies on a circle of radius $1$.
2
Substitute and rearrange
$0.36 + \sin^2\theta = 1$
$\sin^2\theta = 0.64$
3
Take the root, choosing the sign by quadrant
$\sin\theta = \pm 0.8$
In the first quadrant both coordinates are positive, so $\sin\theta = 0.8$. The negative root would correspond to a point below the axis, in the fourth quadrant.
4
State the point and check
$P = (0.6, \ 0.8)$
Checking the distance from the origin: $\sqrt{0.36 + 0.64} = \sqrt{1} = 1$, so the point really is on the unit circle. Both coordinates are within $-1$ and $1$, as required.
Nice work, XP earned
Answer$\sin\theta = 0.8$, and $P = (0.6, 0.8)$
Watch Me Solve It · Consistency with the old definition
+15 XP per step
Q3
PROBLEM
A right-angled triangle has hypotenuse $5$, with an angle $\theta$ whose opposite side is $3$ and adjacent side is $4$. (a) Find $\sin\theta$ and $\cos\theta$ using the old definition. (b) Find the coordinates of the corresponding point on the unit circle, and confirm the two methods agree.
Opposite over hypotenuse, and adjacent over hypotenuse.
2
(b) Scale the triangle to hypotenuse 1
$\text{divide every side by } 5$
The angle is unchanged by scaling, since scaling produces a similar triangle. The sides become $0.6$, $0.8$ and $1$.
3
(b) Read the coordinates
$P = (0.8, \ 0.6)$
The adjacent side, now $0.8$, is the horizontal displacement; the opposite side, now $0.6$, is the vertical one.
4
Compare the two answers
The unit circle gives $\cos\theta = 0.8$ and $\sin\theta = 0.6$, which are exactly the values from part (a). The two definitions agree, as they must for an acute angle. Note that scaling to hypotenuse $1$ is precisely what makes the denominators vanish.
Nice work, XP earned
AnswerBoth methods give $\sin\theta = 0.6$ and $\cos\theta = 0.8$; the point is $(0.8, 0.6)$
Brain Trainer · 5 problems
D
Brain Trainer · Around the circle
5 problems
Five items on the unit circle definition. Work each one, then reveal the answer.
1 State the coordinates of the point at $\theta = 0°$.
The starting point of the turn.$(1, 0)$
2 Find $\cos 180°$.
Half a turn lands at $(-1, 0)$; take the first coordinate.$-1$
3 Find $\sin 270°$.
Three quarters of a turn lands at $(0,-1)$; take the second coordinate.$-1$
4 Can $\sin\theta$ equal $1.4$?
It is a coordinate on a circle of radius $1$.No
5 If $\sin\theta = \tfrac{5}{13}$ and $\theta$ is acute, find $\cos\theta$.
Use $\cos^2\theta = 1 - \tfrac{25}{169}$.$\tfrac{12}{13}$
Complete in your workbook.
Multiple Choice · 5 questions
MC1
The definition
+10 XP
On the unit circle, the point reached by turning through $\theta$ has coordinates:
Correct, cosine is the first coordinate and sine the second, in alphabetical order.
Option A swaps them. Check against $\theta = 0°$: the point is $(1,0)$ and $\cos 0° = 1$, so cosine must come first.
Explanation: Testing a definition at a known angle is the quickest way to settle which way round it goes.
MC2
Why radius one
+10 XP
The circle is chosen to have radius $1$ because:
Correct, with hypotenuse $1$ the denominators in the SOH-CAH-TOA fractions become $1$, so each ratio equals a coordinate directly.
Option D is false and worth being sure about: scaling a triangle produces a similar triangle, which has the same angles and the same ratios.
Explanation: A circle of radius $r$ would give coordinates $(r\cos\theta, r\sin\theta)$, which works but carries an $r$ through every calculation.
MC3
The range
+10 XP
Which of these could be a value of $\cos\theta$?
Correct, cosine is a coordinate of a point on a circle of radius $1$, so it lies between $-1$ and $1$, and $-0.9$ does.
Options A, C and D all lie outside that interval. No angle produces them, so any calculation returning one of them contains an error.
Explanation: This makes a fast check on every answer: if a sine or cosine comes out beyond $\pm 1$, look for the mistake rather than the interpretation.
MC4
Quadrant values
+10 XP
The value of $\cos 90°$ is:
Correct, a quarter turn lands at $(0,1)$, and cosine is the first coordinate, which is $0$.
Option A reads the second coordinate instead of the first; that value is $\sin 90°$.
Explanation: Locate the point first, then read off the coordinate the question asks for. Doing it in that order avoids the swap.
MC5
The identity
+10 XP
The relationship $\sin^2\theta + \cos^2\theta = 1$ holds:
Correct, the point $(\cos\theta, \sin\theta)$ is on the unit circle for every angle, and the identity is just Pythagoras applied to that point.
Option A imposes the restriction the unit circle exists to remove. The identity is if anything more general than the definitions it came from.
Explanation: Squaring removes any sign issues, so the identity is unaffected by which quadrant the point is in.
Short Answer · 3 questions
Q6
Establish the definition
+15 XP
Q6
SHORT ANSWER
(a) State the unit circle definitions of $\sin\theta$ and $\cos\theta$, including the conventions on where the angle is measured from and in which direction. (b) Show that for an acute angle these agree with the right-triangle definitions, explaining what role the radius of $1$ plays. (c) Explain why the unit circle definition applies to $\theta = 130°$ but the right-triangle definition does not. (d) State the coordinates of the points at $0°$, $90°$, $180°$ and $270°$, and hence the sine and cosine at each.
Write your working in your book.
(a) Take the circle of radius $1$ centred at the origin. Starting at $(1,0)$, turn anticlockwise through $\theta$ to reach a point $P$ on the circle. Then $\cos\theta$ is the first coordinate of $P$ and $\sin\theta$ is the second, so $P = (\cos\theta, \sin\theta)$. Angles are measured from the positive horizontal axis, and anticlockwise is the positive direction.
(b) For acute $\theta$, drop a perpendicular from $P$ to the horizontal axis. This gives a right-angled triangle whose hypotenuse is the radius, of length $1$, whose adjacent side has length equal to the first coordinate of $P$, and whose opposite side has length equal to the second. Then $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{\text{first coordinate}}{1}$, which is the first coordinate, and similarly for sine. The radius of $1$ is what makes each denominator equal to $1$, so the ratios reduce to the coordinates themselves; on a circle of radius $r$ the coordinates would be $r\cos\theta$ and $r\sin\theta$ instead.
(c) A right-angled triangle has one angle of $90°$, and its remaining two angles sum to $90°$, so each of them is acute. There is therefore no right-angled triangle containing an angle of $130°$, and the ratio definitions have nothing to refer to. The unit circle definition has no such problem: turning through $130°$ from $(1,0)$ reaches a perfectly ordinary point in the second quadrant, and that point has coordinates like any other.
(d) At $0°$: $(1,0)$, so $\cos 0° = 1$ and $\sin 0° = 0$. At $90°$: $(0,1)$, so $\cos 90° = 0$ and $\sin 90° = 1$. At $180°$: $(-1,0)$, so $\cos 180° = -1$ and $\sin 180° = 0$. At $270°$: $(0,-1)$, so $\cos 270° = 0$ and $\sin 270° = -1$.
Marking guidance: 2 marks for (a), one of which is for both conventions. 3 marks for (b), including the role of the radius. 2 marks for (c). 2 marks for (d).
Q7
Using the identity
+15 XP
Q7
SHORT ANSWER
(a) Derive $\sin^2\theta + \cos^2\theta = 1$ from the unit circle, stating the theorem you use. (b) Given that $\theta$ is in the first quadrant with $\sin\theta = \tfrac{7}{25}$, find $\cos\theta$ exactly. (c) State the coordinates of the corresponding point and verify it lies on the unit circle. (d) Explain why the working in (b) produced two possible values and how you chose between them.
Write your working in your book.
(a) The point $P = (\cos\theta, \sin\theta)$ lies on the unit circle, so its distance from the origin is $1$. By Pythagoras applied to the right triangle formed by $P$, the origin and the foot of the perpendicular from $P$, the square of that distance is the sum of the squares of the coordinates. So $(\cos\theta)^2 + (\sin\theta)^2 = 1^2 = 1$, which is the identity.
(b) Substituting into the identity, $\left(\tfrac{7}{25}\right)^2 + \cos^2\theta = 1$, so $\cos^2\theta = 1 - \tfrac{49}{625} = \tfrac{576}{625}$. Taking the square root, $\cos\theta = \pm\tfrac{24}{25}$, and since $\theta$ is in the first quadrant both coordinates are positive, so $\cos\theta = \tfrac{24}{25}$.
(c) The point is $\left(\tfrac{24}{25}, \tfrac{7}{25}\right)$. Its distance from the origin squared is $\tfrac{576}{625} + \tfrac{49}{625} = \tfrac{625}{625} = 1$, so the distance is $1$ and the point lies on the unit circle as required.
(d) Squaring destroys sign information, so undoing it with a square root always admits both a positive and a negative value: the equation $\cos^2\theta = \tfrac{576}{625}$ genuinely has two solutions. The identity alone cannot distinguish them, because it is unchanged if either coordinate has its sign flipped. The extra information needed is the quadrant, which fixes the signs of both coordinates. Here the first quadrant means both are positive, so the positive root is correct; had $\theta$ been in the second quadrant, the same sine value would have gone with $\cos\theta = -\tfrac{24}{25}$.
Marking guidance: 2 marks for (a), one for naming Pythagoras. 2 marks for (b). 2 marks for (c). 3 marks for (d): one for identifying squaring as the cause, one for why the identity cannot decide, one for the role of the quadrant.
Q8
Reasoning about the circle
+15 XP
Q8
SHORT ANSWER
(a) Explain why turning through $360°$ returns to the same point, and state what this implies about $\sin 400°$ compared with $\sin 40°$. (b) Explain why $\cos\theta$ can equal $0$ for some angles but $\sin^2\theta + \cos^2\theta$ can never equal $0$. (c) Two different angles between $0°$ and $360°$ have the same sine. Using the circle, describe where their points sit relative to each other, and give an example. (d) Explain why no angle has both $\sin\theta = 0.8$ and $\cos\theta = 0.8$.
Write your working in your book.
(a) A full turn is $360°$, so turning through that amount brings you all the way around the circle back to where you began. Turning through $400°$ is therefore a full turn plus a further $40°$, and lands on exactly the same point as turning through $40°$ alone. Since sine is defined as a coordinate of that point, $\sin 400° = \sin 40°$. The same holds for cosine, and it is the first appearance of the repeating behaviour that Lesson 4 makes into a graph.
(b) $\cos\theta = 0$ whenever the point sits on the vertical axis, which happens at $90°$ and at $270°$. That is perfectly possible, since a point on the circle may have first coordinate zero. But $\sin^2\theta + \cos^2\theta$ equals $1$ for every angle, by the identity, and $1 \neq 0$. Geometrically, the sum being zero would require both coordinates to be zero, that is the point to be at the origin, and the origin is not on a circle of radius $1$.
(c) Sine is the second coordinate, so two angles have the same sine exactly when their points sit at the same height. On a circle, two distinct points at the same height are mirror images of each other in the vertical axis: one in the right half and one in the left half. For example $30°$ and $150°$ both give a second coordinate of $0.5$, so $\sin 30° = \sin 150° = 0.5$, while their first coordinates are opposite in sign. This is the supplementary-angle relationship that Lesson 5 establishes in general.
(d) If both were $0.8$, the point would be $(0.8, 0.8)$, and its distance from the origin squared would be $0.64 + 0.64 = 1.28$, giving a distance of about $1.13$. That is not $1$, so the point is not on the unit circle and cannot correspond to any angle. Equivalently, the identity would require $0.64 + 0.64 = 1$, which is false. The largest value both can share is when $\sin\theta = \cos\theta$, which needs $2\sin^2\theta = 1$, so each equals $\tfrac{1}{\sqrt{2}} \approx 0.707$, at $\theta = 45°$.
Marking guidance: 2 marks for (a). 2 marks for (b). 3 marks for (c): one for the reasoning about height, one for the mirror description, one for a correct example. 2 marks for (d).
S
Stretch Challenge · What the circle definition buys
+25 XP
S
CHALLENGE
(a) A point on a circle of radius $r$ centred at the origin is reached by turning through $\theta$. Find its coordinates in terms of $r$ and $\theta$, and explain why the unit circle is the convenient choice. (b) Explain why sine and cosine are functions of the angle, in the sense of the Functions focus area, and state their domain and range. (c) The point $P$ moves anticlockwise around the unit circle at a steady rate. Describe how its height above the horizontal axis changes over one full turn, using the vocabulary of rates of change.
(a) Scaling the unit circle by a factor of $r$ multiplies both coordinates by $r$ without changing any angle, since the result is a similar figure. So the point is $(r\cos\theta, \ r\sin\theta)$. The unit circle is convenient because setting $r = 1$ removes that factor entirely, leaving the coordinates equal to the trigonometric values themselves rather than to $r$ times them. Every later formula is then free of a constant that would otherwise have to be carried through and cancelled.
(b) Each angle determines exactly one point on the circle, and each point has exactly one first coordinate and one second coordinate. So each angle produces exactly one value of $\cos\theta$ and one of $\sin\theta$, which is precisely the condition for a function. Their domain is all real numbers: every angle is allowed, including angles beyond $360°$ and negative ones, since turning further or turning the other way still lands somewhere on the circle. Their range is $[-1, 1]$ for both, since a coordinate of a point on a circle of radius $1$ cannot exceed $1$ in size, and every value in that interval is attained by some point. Note that these are many-to-one functions: many different angles share a sine, as part (c) of the previous question showed, which is fine for a function and is why solving a trigonometric equation gives more than one answer.
(c) The height is the second coordinate, which is $\sin\theta$. Starting at $\theta = 0°$ the point is at height $0$ and moving upward fastest, since near the horizontal axis almost all of its motion is vertical. As it approaches $90°$ it is still rising but more and more slowly, because near the top of the circle its motion is almost entirely sideways: so from $0°$ to $90°$ the height is increasing at a decreasing rate, reaching a maximum of $1$. From $90°$ to $180°$ it falls, slowly at first and then faster, so it is decreasing at an increasing rate, returning to $0$. From $180°$ to $270°$ it continues down to $-1$, decreasing at a decreasing rate, and from $270°$ to $360°$ it climbs back to $0$, increasing at an increasing rate. All four of the qualitative shapes from the Rates of Change focus area appear in one turn, which is exactly why the graph of $\sin\theta$ has the smooth wave shape drawn in Lesson 4.
R
Quick Review
recap
The definition
$P = (\cos\theta, \sin\theta)$ on the unit circle
Conventions
From the positive horizontal axis, anticlockwise
It extends
Agrees with SOH-CAH-TOA for acute angles
Free facts
Range $[-1,1]$, and $\sin^2\theta + \cos^2\theta = 1$
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Mark lesson as complete
Tick when you've finished Learn, Practice and the Stretch. Earns +90 XP and +25 coins.