3.8 The unit circle and trigonometric equations

Define sine and cosine for any angle on the unit circle, use tan θ = sin θ/cos θ and cos²θ + sin²θ = 1, handle the ambiguous case of the sine rule, and solve trigonometric equations graphically.

IB Mathematics: Applications and Interpretation
Part of 3 Geometry and trigonometry
HL only

Key terms introduced here

Each is in the Maths AI glossary, with every other term in the subject.

Ambiguous case of the sine rule: The situation in which two sides of a triangle and an angle that is not between them are given, and the data fit two different triangles, one with an acute angle and one with the obtuse supplement of that angle.
Cosine function: The function whose value $\cos\theta$ is the $x$ -coordinate of the point where the arm of the angle $\theta$ , in standard position, meets the unit circle.
Periodic function: A function whose values repeat at regular intervals, so that $f(x + p) = f(x)$

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3.8 The unit circle and trigonometric equations · IB Maths AI | Augment Education

3.8 The unit circle and trigonometric equations

IB Mathematics: Applications and Interpretation
Part of 3 Geometry and trigonometry
HL only

Key terms introduced here

Each is in the Maths AI glossary, with every other term in the subject.

Ambiguous case of the sine rule: The situation in which two sides of a triangle and an angle that is not between them are given, and the data fit two different triangles, one with an acute angle and one with the obtuse supplement of that angle.
Cosine function: The function whose value $\cos\theta$ is the $x$ -coordinate of the point where the arm of the angle $\theta$ , in standard position, meets the unit circle.
Periodic function: A function whose values repeat at regular intervals, so that $f(x + p) = f(x)$

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Notes on 3.8 The unit circle and trigonometric equations

35 min readLast reviewed 15 Oct 2026

Learning objectives

State the definitions of $\cos\theta$ and $\sin\theta$ as the coordinates of a point on the unit circle, for any angle $\theta$ .
Explain how the graphs of $y = \sin x$ and $y = \cos x$ are built from the unit circle, and write down their key features.
Determine the sign of $\sin\theta$ , $\cos\theta$ and $\tan\theta$ in each quadrant, and use the symmetries of the unit circle to relate angles.
Use the definition $\tan\theta = \frac{\sin\theta}{\cos\theta}$ .
Use the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$ to find one trigonometric value from another.
Determine whether two sides and a non-included angle give no triangle, one triangle or two triangles, and find every possible triangle with the sine rule or the cosine rule.
Solve trigonometric equations graphically in a finite interval using technology, finding every solution in the interval.
Interpret the solutions of a trigonometric equation in the context of a rotating or repeating situation.

Introduction

In a right-angled triangle, sine, cosine and tangent are ratios of sides, so they only make sense for angles between 0° and 90°. Rotating objects, waves, tides and alternating currents turn through angles of any size, many times over, and their mathematics needs a sine and cosine for every angle. The unit circle provides them by taking $\cos\theta$ and $\sin\theta$ to be the coordinates of a point on a circle of radius 1, whatever the angle.

This one definition explains why the sine rule can give two different triangles from the same data, why the graphs of sine and cosine repeat every $2\pi$ , and why a trigonometric equation such as $\sin x = 0.4$ has many solutions. Equations of this kind arise whenever the time at which a rotating or oscillating quantity reaches a particular value is needed.

The unit circle

The definitions start with a circle of radius 1, because then the hypotenuse of every right-angled triangle drawn from its centre is 1, and the ratios become plain lengths.

Every angle is measured from the same starting line and in the same direction, so that each angle corresponds to exactly one point on the circle.

Cosine and sine as coordinates

Draw an angle $\theta$ in standard position and let its second arm meet the unit circle at the point P. For an acute angle, dropping a perpendicular from P to the $x$ -axis makes a right-angled triangle with hypotenuse 1. By the cosine ratio the horizontal side is $1 \times \cos\theta = \cos\theta$ , and by the sine ratio the vertical side is $1 \times \sin\theta = \sin\theta$

Unit circle centre O on x- and y-axes through (1, 0), (0, 1), (−1, 0), (0, −1). Radius of length 1 at angle θ to the positive x-axis meets the circle at P(cos θ, sin θ); a right-angled triangle has horizontal side cos θ and vertical side sin θ. — For an acute angle, the right-angled triangle inside the unit circle has hypotenuse 1, so its horizontal and vertical sides are cos θ and sin θ: the coordinates of P.

The unit circle figure shows this triangle. For acute angles the coordinates and the triangle ratios agree. For every other angle there is no triangle to start from, so the coordinates themselves become the definition.

Extending a definition in this way is an example of generalization: the new definition agrees with the old one wherever the old one made sense (acute angles), and also gives a value wherever the old one did not (0, obtuse angles, reflex angles, negative angles and angles of more than a full turn). The four points where the circle meets the axes give the values at quarter-turns directly.

$\theta$	0	$\frac{\pi}{2}$ (90°)	$\pi$ (180°)	$\frac{3\pi}{2}$

Because every point on the unit circle has both coordinates between −1 and 1, sine and cosine always take values in the interval $-1 \le y \le 1$ . An equation such as $\sin\theta = 1.2$ therefore has no solution.

Quadrants, signs and symmetry

The four quadrants

The axes divide the plane into four regions, numbered anticlockwise from the top right. The sign of each coordinate, and so of each trigonometric value, depends only on which region the arm of the angle lies in.

Unit circle with four symmetric points at angles θ, π − θ, π + θ and 2π − θ joined by dashed lines. Signs: quadrant 1 sin +, cos +, tan +; quadrant 2 sin +, cos −, tan −; quadrant 3 sin −, cos −, tan +; quadrant 4 sin −, cos +, tan −. — The four points at θ, π − θ, π + θ and 2π − θ are reflections of one another, so their sines and cosines have the same size and differ only in sign.

The quadrant diagram shows the signs in each region. In quadrant 1 all three functions are positive; in quadrant 2 only the sine; in quadrant 3 only the tangent; in quadrant 4 only the cosine. Some books remember this as "CAST", reading the positive function in each quadrant anticlockwise from quadrant 4: Cosine, All, Sine, Tangent.

Quadrant	Angles (radians)	Angles (degrees)	$\sin\theta$	$\cos\theta$	$\tan\theta$
1	$0 < \theta < \frac{\pi}{2}$

Symmetries of the unit circle

The quadrant diagram also shows that the four points at $\theta$ , $\pi - \theta$ , $\pi + \theta$ and $2\pi - \theta$ are mirror images of one another in the axes. Reflecting the point

\begin{aligned} \sin(\pi - \theta) &= \sin\theta, & \cos(\pi - \theta) &= -\cos\theta \\ \sin(\pi + \theta) &= -\sin\theta, & \cos(\pi + \theta) &= -\cos\theta \\ \sin(-\theta) &= -\sin\theta, & \cos(-\theta) &= \cos\theta \end{aligned}

The angle $-\theta$ is measured clockwise, so it ends at the same point as $2\pi - \theta$ . The first line matters most in practice, because it is the reason the sine rule can give two different triangles from the same data.

In degrees, $\sin(180^\circ - \theta) = \sin\theta$ says that an acute angle and the obtuse angle that makes it up to 180° share the same sine. The inverse sine on a GDC only ever returns the acute one.

Angles beyond a full turn

An angle of $2\pi + \theta$ is a full turn followed by $\theta$ , so it ends at exactly the same point as $\theta$ . Any whole number of extra turns, in either direction, makes no difference to the point, and so no difference to the sine or the cosine.

\sin(\theta + 2k\pi) = \sin\theta, \qquad \cos(\theta + 2k\pi) = \cos\theta \qquad (k \in \mathbb{Z})

The smallest such $p$ is the period of the function, the length of one complete cycle. Periodicity is the reason trigonometric functions model repeating situations: the height of a point on a turning wheel, the depth of water in a tidal harbour, the voltage of an alternating current.

The graphs of sine and cosine

As the angle $\theta$ increases from 0 to $2\pi$ , the point P travels once round the unit circle anticlockwise. Plotting its $y$ -coordinate against the angle gives the graph of $y = \sin\theta$ ; plotting its $x$ -coordinate gives $y = \cos\theta$

From 0 to $\frac{\pi}{2}$ , P climbs from $(1, 0)$ to $(0, 1)$ : the sine rises from 0 to 1 and the cosine falls from 1 to 0.
From $\frac{\pi}{2}$

After $2\pi$ the point retraces the same path, so the graphs repeat.

The graph of sine and cosine shows two curves of exactly the same shape. The cosine curve is the sine curve shifted $\frac{\pi}{2}$ to the left, because the $x$ -coordinate of a point at angle $\theta$ equals the $y$ -coordinate of a point a quarter-turn further round: $\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right)$

Feature	$y = \sin x$	$y = \cos x$
Value at $x = 0$	0	1
Maximum value	1, at $x = \frac{\pi}{2}$

These two curves are the building blocks of every sinusoidal model in subtopic 2.5: stretching them vertically changes the amplitude, stretching them horizontally changes the period, and moving them up or down changes the principal axis.

Khan Academy, “Unit circle definition of trig functions”: watch how the coordinates of a point on the unit circle give the sine and cosine of any angle, including obtuse and negative ones.

The Khan Academy video works through the same definition as the unit circle figure, showing the coordinates of the point changing sign as the angle passes into the second, third and fourth quadrants.

The tangent function

In a right-angled triangle, $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ . In the unit-circle triangle the opposite side is $\sin\theta$ and the adjacent side is $\cos\theta$

The tangent has a geometric meaning on the unit circle. The radius OP runs from $(0, 0)$ to $(\cos\theta, \sin\theta)$ , so its gradient is $\frac{\sin\theta - 0}{\cos\theta - 0} = \tan\theta$

The sign of $\tan\theta$ follows from the signs of sine and cosine: positive when they agree (quadrants 1 and 3) and negative when they differ (quadrants 2 and 4). The arms at $\theta$ and $\theta + \pi$ point in opposite directions along the same line, so they have the same gradient, and the tangent has period $\pi$ , half the period of sine and cosine.

The tangent graph shows vertical asymptotes where the cosine is zero, because division by zero is undefined there. Between the asymptotes the curve rises through every real value, so an equation such as $\tan x = 50$ has solutions, while $\sin x = 50$ has none.

The Pythagorean identity

The point $(\cos\theta, \sin\theta)$ lies on the unit circle, whose equation is $x^2 + y^2 = 1$ . Substituting the coordinates gives a result true for every angle. Equivalently, for an acute angle, Pythagoras' theorem in the unit-circle triangle with legs $\cos\theta$

\cos^2\theta + \sin^2\theta = 1

Here $\sin^2\theta$ means $(\sin\theta)^2$ , the square of the sine. An equation that holds for every value of its variable has a special name.

An equation such as $\sin\theta = 0.4$ is true only for particular angles, so it is solved; an identity is true for all angles, so it is used to rewrite one expression as another. The identity used most often comes straight from the equation of the unit circle.

Finding one value from another

The identity gives $\cos\theta$ from $\sin\theta$ (or the reverse), but only up to its sign, because squaring loses the sign. The quadrant of the angle decides which sign is correct.

Exact values (for understanding only)

Two triangles give exact values for the angles $\frac{\pi}{6}$ , $\frac{\pi}{4}$ and $\frac{\pi}{3}$

Table of exact values

$\theta$	$\frac{\pi}{6}$ (30°)	$\frac{\pi}{4}$

The ambiguous case of the sine rule

The sine rule of subtopic 3.2 finds an angle from the sine of that angle. Because $\sin\theta = \sin(180^\circ - \theta)$ , every sine between 0 and 1 belongs to two angles in a triangle's range of 0° to 180°: an acute one and its obtuse supplement. The GDC's $\sin^{-1}$ gives only the acute angle. When the obtuse angle also fits in a triangle with the given data, there are two different triangles, and both must be found.

Why two triangles can exist

Suppose angle $A$ and the side $b = \text{AC}$ next to it are fixed, and the side $a = \text{BC}$ opposite $A$ has a given length. The vertex C is fixed, and B must lie on the base line from A at a distance $a$ from C. The possible positions of B are the points where a circle of radius

Triangle with angle A = 35° and side b = AC = 11. A dashed arc of radius a = 8 centred at C cuts the base line from A at two points, B₁ and B₂, giving two triangles AB₁C and AB₂C. A dashed perpendicular h from C to the base. — The side a = 8 can swing to meet the base line at B₁ or B₂, because it is longer than the height h = 11 sin 35° ≈ 6.31 but shorter than b = 11. The angle at B₁ is the supplement of the angle at B₂.

The ambiguous-case figure shows the arc of radius 8 crossing the base line at two points. Triangle CB₁B₂ is isosceles (both sides from C are 8), so its base angles are equal, and the angle $\text{A}\hat{\text{B}}_1\text{C}$ is the supplement of that base angle. That is the geometric reason the two angles at B are $\theta$ and $180^\circ - \theta$

When there are no, one or two triangles

The number of triangles depends on how the length $a$ compares with the perpendicular height from C to the base line, $h = b\sin A$ , and with $b$ .

Given angle $A$	Condition on $a$	Number of triangles	Why
acute	$a < b\sin A$	0	the side is too short to reach the base line
acute	$a = b\sin A$

In practice, a calculation with the sine rule shows which case applies without the table:

Find $\sin B = \frac{b\sin A}{a}$ . If this is greater than 1, there is no triangle (the GDC gives an error for $\sin^{-1}$

The cosine-rule method

The cosine rule gives both triangles at once, as the two roots of a quadratic. Taking the unknown side as $x$ and writing the cosine rule for the given angle gives an equation of the form $x^2 - (2b\cos A)x + (b^2 - a^2) = 0$

Ambiguous triangles in radians

The same reasoning works in radians, with $\pi - \theta$ as the second angle: the GDC must then be in radian mode, and the angle sum of a triangle is $\pi$ .

Solving trigonometric equations graphically

An equation in which the unknown is an angle inside a trigonometric function is solved for the angles that make it true.

Why there is more than one solution

The equation $\sin x = 0.4$ asks for the angles whose unit-circle point is at height 0.4. The horizontal line $y = 0.4$ crosses the unit circle twice, at $x = \sin^{-1}(0.4)$

The graph of $y = \sin x$ and $y = 0.4$ shows both solutions in $0 \le x \le 2\pi$ and their symmetry about the peak at $\frac{\pi}{2}$

The graphical method with technology

Most trigonometric equations in applications cannot be rearranged neatly, so they are solved by graphing. The method has the same steps every time.

Check the angle unit and set the GDC to radian or degree mode.
Enter the left side as $Y_1$ and the right side as $Y_2$ (or rearrange to $f(x) = 0$

This is the graphical solution method of subtopic 2.4, applied to periodic functions. The number of solutions is visible on the screen, so the method finds solutions that the inverse function on its own would miss.

Many equations mix trigonometric and other functions, cannot be solved by algebra at all, and are solved by graphing.

Equations in context

When a trigonometric function models a rotating or repeating quantity, solving an equation answers the question "when does the quantity reach this value?". The interval comes from the time period in the question, and every solution in it is a moment when the value is reached.

The graph of the bucket's height shows each answer: the four crossings of $h = 4$ , the two 10-second stretches above that line, and the short dips below the water level at the start and end of each revolution. Reading the graph before calculating helps to avoid missing a solution or giving one outside the interval.

Eddie Woo, “Sine Rule: The Ambiguous Case”: watch how a side of fixed length swings to meet the base line at two points, and why the two angles found are supplementary.

The Eddie Woo video works through the ambiguous case with the same swinging-side construction as the ambiguous-case figure, and is a useful review of the conditions for two triangles before attempting mixed questions.

Exam focus

The unit circle and trigonometric equations are examined on HL Papers 1 and 2, usually inside larger questions. The ambiguous case appears as a triangle or enclosure problem: show that two triangles exist, find both possible angles or sides (by the sine rule with $180^\circ - \theta$ , or by solving the cosine-rule quadratic), and then find or compare their areas, often choosing the triangle that fits a condition. Graphical equation solving appears inside sinusoidal-model questions from subtopics 2.5 and 2.9: find the times when a height or depth reaches a given value, or the total time it stays above a level. Identities are used to find $\cos\theta$ or $\tan\theta$

The most common lost marks are a missing obtuse angle, a GDC in degree mode for an angle given in radians, and solutions outside the interval. In (b), giving only 1.01 loses an A1 and every later mark that needs the second triangle.

When a question asks you to choose between two triangles (the larger area, the shorter side, an obtuse angle), find both first and then state which one is chosen and why. Calculus is not needed to "maximise" the area, because there are only two candidates to compare.

Practice questions

An angle of 5 radians is drawn in standard position. Find the coordinates of the point where its arm meets the unit circle, and state the quadrant. [3]
Find all values of $x$ , $0 \le x \le 2\pi$ , for which $\cos x = -0.3$ . [3]
Given that $\sin\theta = -0.35$

Answers

$(\cos 5, \sin 5) = (0.284, -0.959)$ ; quadrant 4, since $\frac{3\pi}{2} \approx 4.71 < 5 < 2\pi$

Summary

On the unit circle (radius 1, centre the origin), the point at angle $\theta$ in standard position is $(\cos\theta, \sin\theta)$ ; this defines sine and cosine for every angle, and both take values from −1 to 1.
Signs by quadrant: all positive in quadrant 1; only sine in 2; only tangent in 3; only cosine in 4.
Symmetries: $\sin(\pi - \theta) = \sin\theta$

Self-check

To do:
I can state the coordinates of the point on the unit circle at any angle $\theta$ .
To do:
I can find $\sin\theta$ and $\cos\theta$ at 0, $\frac{\pi}{2}$

Questions on 3.8

39 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on 20 October 2026

Question 1 · AI-0015314 marks

The Riverside Wheel is a large observation wheel with radius 40 m. Its centre is 45 m above the ground. The wheel turns anticlockwise at a constant speed and completes one rotation every 15 minutes. There are 24 passenger capsules, equally spaced around the rim.

(a)
Write down, in radians, the angle at the centre of the wheel between two adjacent capsules. Give your answer as an exact multiple of $\pi$ .
1 mark
Markscheme (a)
$\frac{\pi}{12}$ A1
[1 mark]
(b)(i)
Find the distance between two adjacent capsules, measured along the rim.
2 marks
Markscheme (b)(i)
$40 \times \frac{\pi}{12}$ (M1)
$10.5\text{ m}$ $\left(\frac{10\pi}{3},\ 10.4719\ldots\right)$
(b)(ii)
Find the straight-line distance between two adjacent capsules.
2 marks
Markscheme (b)(ii)
EITHER
$\sqrt{40^2 + 40^2 - 2(40)(40)\cos\frac{\pi}{12}}$
(c)
Write down the angular speed of the wheel in radians per minute.
1 mark
Markscheme (c)
$\frac{2\pi}{15}$ $(0.419)$ A1
[1 mark]
(d)
Find the distance travelled by a capsule in 4 minutes.
2 marks
Markscheme (d)
$40 \times 4 \times \frac{2\pi}{15}$ (M1)
$67.0\text{ m}$
(e)
Valentina boards a capsule at the lowest point of the wheel at time $t = 0$ . Her height above the ground, $h$ metres, $t$ minutes later is modelled by
$h(t) = 45 - 40\cos\left(\frac{2\pi t}{15}\right).$
(f)
Find the length of time, during one rotation, for which Valentina is more than 70 m above the ground. Give your answer in minutes.
4 marks
Markscheme (f)
$45 - 40\cos\left(\frac{2\pi t}{15}\right) = 70$

Question 2 · AI-0015617 marks

A pendulum consists of a small bob on a light rod of length 1.2 m. The angle, $\theta$ radians, between the rod and the vertical $t$ seconds after the bob is released is modelled by

\theta(t) = 0.3\cos(2.5t), \quad t \ge 0.

Question 3 · AI-0042220 marks

A car is tested on an elliptical track. Its position vector, in metres, relative to the centre of the track, $t$ seconds after the test begins is

\boldsymbol{r} = \begin{pmatrix} 400\cos(0.05t) \\ 250\sin(0.05t) \end{pmatrix}.

Key terms introduced here

Carry on with 3.8 in Augment

Key terms introduced here

Carry on with 3.8 in Augment

Notes on 3.8 The unit circle and trigonometric equations

Learning objectives

Introduction

The unit circle

Cosine and sine as coordinates

Quadrants, signs and symmetry

The four quadrants

Symmetries of the unit circle

Angles beyond a full turn

The graphs of sine and cosine

The tangent function

The Pythagorean identity

Finding one value from another

Exact values (for understanding only)

The ambiguous case of the sine rule

Why two triangles can exist

When there are no, one or two triangles

The cosine-rule method

Ambiguous triangles in radians

Solving trigonometric equations graphically

Why there is more than one solution

The graphical method with technology

Equations in context

Exam focus

Practice questions

Summary

Self-check

Questions on 3.8

Markscheme (a)

Markscheme (b)(i)

Markscheme (b)(ii)

Markscheme (c)

Markscheme (d)

Markscheme (f)

From the formula booklet

Markscheme (e)