IB Maths AA SL Topic 3 — Geometry & Trig Paper 1 & 2 ~10 min read

The Unit Circle

The unit circle is a circle of radius 1 centred at the origin. It looks simple, but it’s the secret weapon for understanding sin, cos, and tan for any angle — not just the ones in right-angled triangles. Master it, and trig equations stop being scary.

📘 What you need to know

What is the unit circle?

The unit circle is just a circle with two special features: its radius is 1 and its centre is at the origin (0, 0). Pick any point on it and label it (x, y). Now draw a line from the origin to that point — that radius makes an angle θ with the positive x-axis, measured anticlockwise.

The unit circle — radius 1, centre (0, 0)
x y (1, 0) (−1, 0) (0, 1) (0, −1) P(cos θ, sin θ) cos θ sin θ θ O

Sin, cos, and tan from the unit circle

This is the big idea: pick any point on the unit circle. Its coordinates ARE the cos and sin of the angle. Here’s how to remember it:

Cosine
cos θ = x
The x-coordinate of the point
Sine
sin θ = y
The y-coordinate of the point
Tangent
tan θ = yx
Gradient of the line from origin to point

🤔 Why do these definitions match SOH CAH TOA?

Drop a perpendicular from the point on the circle down to the x-axis. You’ve made a right-angled triangle with hypotenuse 1 (the radius). The opposite side is y, the adjacent side is x. So sin θ = opp/hyp = y/1 = y, and cos θ = adj/hyp = x/1 = x. The unit circle just extends these definitions to angles bigger than 90° — which SOH CAH TOA can’t handle on its own.

The unit circle is what makes sin and cos work for angles like 150°, 270°, or even −45°. SOH CAH TOA only works inside right-angled triangles (so 0°–90°). The unit circle lets us go all the way around.

The four quadrants — when is each ratio positive?

Split the circle into 4 quadrants by the axes. In each quadrant, only certain ratios are positive (the rest are negative). The pattern is captured by the famous CAST rule:

S
Quadrant 2
90° to 180°
Only sin positive
A
Quadrant 1
0° to 90°
All positive
T
Quadrant 3
180° to 270°
Only tan positive
C
Quadrant 4
270° to 360°
Only cos positive

🧠 Memory trick — “All Students Take Calculus”

Starting from quadrant 1 and going anticlockwise: All, Students, Take, Calculus. The letter tells you which ratio is positive in that quadrant — everything else is negative. Quadrant 1 = All positive. Quadrant 2 = Sin only. Quadrant 3 = Tan only. Quadrant 4 = Cos only.

The CAST rule on the unit circle
A All S Sin only T Tan only C Cos only 90° 180° 270°

Symmetries — using the unit circle for any angle

The unit circle has obvious symmetries: it’s symmetric across both axes and the origin. This means there are simple relationships between the trig values of related angles.

Angle
sin
cos
tan
−θ
−sin θ
cos θ
−tan θ
180 − θ
sin θ
−cos θ
−tan θ
180 + θ
−sin θ
−cos θ
tan θ
360 − θ
−sin θ
cos θ
−tan θ
360 + θ
sin θ
cos θ
tan θ
The shortcut: identify which quadrant the angle lives in, then use CAST to decide the sign. The “size” of the answer comes from the related acute angle (called the reference angle).
For example, sin 210° lives in quadrant 3, where only tan is positive — so sin is negative. The reference angle is 210 − 180 = 30°. So sin 210° = −sin 30° = −½.

From the unit circle to the sin and cos graphs

If you walk around the unit circle and plot the y-coordinate at each angle, you trace out the sine graph. Plot the x-coordinate instead and you get the cosine graph. That’s where the wave shape comes from.

Sine — from y-coordinate

sin 0 = 0, sin π2 = 1

Starts at 0, peaks at 1, back to 0, down to −1, back to 0. Repeats every 2π.

Cosine — from x-coordinate

cos 0 = 1, cos π2 = 0

Starts at 1, drops to 0, down to −1, back to 0, returns to 1. Repeats every 2π.

Using the unit circle to solve trig equations

The big payoff: when you solve sin θ = 0.5 (or any trig equation), there are usually two answers in 0°–360° — not one. The unit circle shows you exactly where they are.

Method — finding all solutions in the interval 1. Use sin⁻¹/cos⁻¹/tan⁻¹ to find the principal value
2. Mark this point on the unit circle
3. Find the second point giving the same trig value
4. Read off the second angle using symmetries
5. Add/subtract 360° (or 2π) to find more solutions in the interval
The pattern for finding the second solution:
sin θ = k → second point has the same y-coordinate (mirrored across the y-axis)
cos θ = k → second point has the same x-coordinate (mirrored across the x-axis)
tan θ = k → second point is the diagonal opposite (180° away)

Worked examples

WE 1

Find the angle from a point on the circle (SME-style)

The coordinates of a point on a unit circle, to 3 s.f., are (0.629, 0.777). The radius from the centre to the point makes an angle θ° with the positive x-axis. Find θ° to the nearest degree.

Step 1: Identify cos and sin Point on the circle is (cos θ, sin θ). cos θ = 0.629, sin θ = 0.777 Step 2: Use either ratio with inverse trig θ = cos⁻¹(0.629) = 51.023… θ ≈ 51° (nearest degree) both x and y are positive → quadrant 1 → answer is acute
WE 2

Use CAST to find the sign

Without using a calculator, decide whether each of the following is positive or negative:

(a) sin 200°  (b) cos 320°  (c) tan 130°

Step 1: Identify each quadrant (a) 200° → quadrant 3 (180°–270°) (b) 320° → quadrant 4 (270°–360°) (c) 130° → quadrant 2 (90°–180°) Step 2: Apply CAST (a) Q3 → only tan positive → sin 200° is negative (b) Q4 → only cos positive → cos 320° is positive (c) Q2 → only sin positive → tan 130° is negative (a) −   (b) +   (c) −
WE 3

Use symmetry to relate angles

Given that sin 25° ≈ 0.423, find sin 155° and sin 205° without a calculator.

Step 1: Spot the relationship 155° = 180 − 25 → quadrant 2 (sin positive) 205° = 180 + 25 → quadrant 3 (sin negative) Step 2: Apply symmetries sin(180 − θ) = sin θ → sin 155° = sin 25° = 0.423 sin(180 + θ) = −sin θ → sin 205° = −sin 25° = −0.423 sin 155° ≈ 0.423,   sin 205° ≈ −0.423 CAST tells you the sign, the reference angle gives the size!
WE 4

Find both solutions of a sine equation

Solve sin θ = 0.5 for 0° ≤ θ ≤ 360°.

Step 1: Find the principal value θ = sin⁻¹(0.5) = 30° Step 2: Find the second solution using sin(180 − θ) = sin θ θ = 180 − 30 = 150° Step 3: Check both fit the interval Both 30° and 150° are between 0° and 360° ✓ θ = 30° or 150° sin = 0.5 → quadrants where sin is positive (Q1 and Q2)
WE 5

Find all solutions in radians (SME-style)

Given that one solution of cos θ = 0.8 is θ = 0.6435 radians (4 d.p.), find all other solutions in the range −2π ≤ θ ≤ 2π. Give answers to 3 s.f.

Step 1: Identify the principal value θ = 0.6435 (in quadrant 1, cos positive) Step 2: Cosine is positive in Q1 and Q4 — second solution Use cos(−θ) = cos θ → second solution = −0.6435. Step 3: All values in the interval are ±0.6435 ± 2π 0.6435, −0.6435 (already in [−2π, 2π]) 0.6435 − 2π ≈ −5.640 −0.6435 + 2π ≈ 5.640 θ ≈ −5.64, −0.644, 0.644, 5.64 cos repeats every 2π — add/subtract until outside the range

💡 Top tips

⚠ Common mistakes

The unit circle isn’t just a definition — it’s the visual tool that ties together sin, cos, tan, their signs, their symmetries, and the solutions to trig equations. Sketch it for every problem and you’ll see what’s going on instantly.

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