IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

The Unit Circle

The unit circle extends sin, cos, and tan to any angle — not just the acute ones inside a right triangle. A point on the circle has coordinates (cos θ, sin θ). That single idea unlocks angles bigger than 90°, negative angles, and the symmetries you’ll use to solve trig equations.

📘 What you need to know

Unit circle: circle of radius 1 centred at the origin (0, 0).
Angle θ: measured from the positive x-axis. Anticlockwise = positive; clockwise = negative.
Coordinates: a point on the circle is (cos θ, sin θ). So cos θ = x, sin θ = y, tan θ = y/x.
Four quadrants — CAST: Q1 All positive, Q2 Sin only, Q3 Tan only, Q4 Cos only. Mnemonic: “All Students Take Calculus”.
Symmetries: sin(180° − θ) = sin θ; cos(360° − θ) = cos θ; etc. Use these to relate any angle to an acute one.
Periodicity: adding 360° (or 2π) to any angle leaves sin, cos, tan unchanged.
Solving sin/cos/tan = k: use inverse trig for the principal value, then the unit circle for the second solution.

Defining sin, cos, and tan with the unit circle

For any angle θ measured anticlockwise from the positive x-axis, draw a radius. The point where it meets the unit circle has coordinates determined entirely by θ:

Unit circle definitions point = (cos θ, sin θ) tan θ = sin θcos θ

This works for any angle: greater than 90°, negative, or beyond 360°. For acute θ, it agrees with SOH CAH TOA from the right-triangle definition.

The four quadrants — CAST

Quadrant	Angles	Positive ratios
Q1	0° – 90°	All: sin, cos, tan
Q2	90° – 180°	Sin only
Q3	180° – 270°	Tan only
Q4	270° – 360°	Cos only

“All Students Take Calculus” — first letter of each word marks the positive ratio in each quadrant going anticlockwise from Q1: All, Sin, Tan, Cos.

Symmetries of the unit circle

Any angle’s trig value can be written in terms of an acute angle’s trig value, with possibly a flipped sign. The pattern depends on which quadrant the angle sits in.

Angle	sin	cos	tan
180° − θ (Q2)	+sin θ	−cos θ	−tan θ
180° + θ (Q3)	−sin θ	−cos θ	+tan θ
360° − θ (Q4)	−sin θ	+cos θ	−tan θ
−θ	−sin θ	+cos θ	−tan θ
360° + θ	+sin θ	+cos θ	+tan θ

Don’t memorise this whole table — just remember CAST and the rule “the magnitude is the same, the sign comes from the quadrant”. Then derive each line on the spot.

Solving simple trig equations

For an equation like sin x = k, the calculator gives one answer (the principal value). The unit circle gives a second answer in the range. Then add or subtract 360° (or 2π) to reach all solutions in the required interval.

sin x = k

x and 180° − x

same y-coordinate (horizontal mirror)

cos x = k

x and 360° − x

same x-coordinate (vertical mirror)

For tan x = k, the second solution is 180° + x (radius along the same line, opposite direction).

🧭 Recipe — solving sin/cos/tan = k

Principal value: x₁ = sin⁻¹(k), cos⁻¹(k), or tan⁻¹(k) on the calculator.
Second value: sin → 180° − x₁; cos → 360° − x₁; tan → 180° + x₁.
Extend to the interval: add or subtract 360° (or 2π) until you’ve covered the whole range.
Check all answers lie inside the required interval.

Worked examples

WE 1

Find an angle from unit-circle coordinates

A point on the unit circle has coordinates (0.6, 0.8). The radius from the centre to the point makes an angle θ with the positive x-axis, where 0° ≤ θ ≤ 360°. Find θ, correct to the nearest degree.

Step 1: Identify cos θ and sin θ from the coordinates cos θ = 0.6, sin θ = 0.8 Step 2: Both positive → Q1 θ = cos⁻¹(0.6) = 53.13° θ ≈ 53° (nearest degree) 0.6² + 0.8² = 1 ✓ confirms the point is on the unit circle (3-4-5 triangle scaled to radius 1)

WE 2

Express trig values in terms of an acute angle

Express sin 200°, cos 200°, and tan 200° in terms of trig values of the acute angle 20°.

Step 1: Identify the quadrant 200° is between 180° and 270° → Q3 In Q3: only tan is positive Step 2: Write 200° = 180° + 20° and apply symmetries sin 200° = sin(180° + 20°) = −sin 20° cos 200° = cos(180° + 20°) = −cos 20° tan 200° = tan(180° + 20°) = +tan 20° −sin 20°, −cos 20°, tan 20° signs match Q3: sin and cos negative, tan positive ✓

WE 3

Find the quadrant from signs

The angle θ satisfies sin θ < 0 and tan θ > 0. State the quadrant in which θ lies.

Step 1: Use CAST to filter quadrants sin θ < 0 → Q3 or Q4 tan θ > 0 → Q1 or Q3 Step 2: Take the intersection both conditions → Q3 θ is in Q3 (180° < θ < 270°)

WE 4

Find missing trig values using a quadrant constraint

Given that sin α = 3/5 and α lies in the second quadrant, find the exact values of cos α and tan α.

Step 1: Use sin²α + cos²α = 1 (Pythagorean identity) cos²α = 1 − (3/5)² = 1 − 9/25 = 16/25 cos α = ±4/5 Step 2: Use the quadrant to fix the sign α in Q2 → cos α negative cos α = −4/5 Step 3: tan α = sin α / cos α tan α = (3/5) ÷ (−4/5) = −3/4 tan α = −3/4 CAST confirms Q2: sin positive, cos and tan negative ✓

WE 5

Solve sin x = 0.5 for 0° ≤ x ≤ 360°

Find all values of x in the interval 0° ≤ x ≤ 360° that satisfy sin x = 0.5.

Step 1: Principal value x₁ = sin⁻¹(0.5) = 30° Step 2: Second solution — sin positive in Q1 and Q2 x₂ = 180° − 30° = 150° Step 3: Both lie in [0°, 360°] — no need to add/subtract 360° x = 30° or x = 150° two solutions per 360° period for sin x = positive constant

WE 6

Solve sin θ = 0.4 for −2π ≤ θ ≤ 2π

Find all values of θ in the interval −2π ≤ θ ≤ 2π that satisfy sin θ = 0.4. Give answers correct to 3 s.f.

Step 1: Principal value (calculator in radian mode) θ₁ = sin⁻¹(0.4) = 0.4115… Step 2: Second solution — sin positive in Q1 and Q2 θ₂ = π − 0.4115… = 2.7300… Step 3: Subtract 2π for the negative half of the interval θ₁ − 2π = 0.4115 − 6.2832 = −5.8717… θ₂ − 2π = 2.7300 − 6.2832 = −3.5532… Step 4: Collect all solutions in [−2π, 2π] θ ≈ −5.87, −3.55, 0.412, 2.73 (3 s.f.) interval of length 4π contains exactly four solutions for sin θ = constant ≠ ±1

💡 Top tips

Sketch the unit circle on every problem. The diagram tells you which quadrant and where the second solution sits.
CAST is the fastest sign-checker. Memorise “All Students Take Calculus” once and you’ll never confuse the signs again.
Calculator mode matters: degree problems → DEG; radian problems → RAD.
Two solutions per period for sin/cos = constant; one per period for tan = constant.
Always include the principal value in your final answer set — it’s easy to forget.

⚠ Common mistakes

Only writing the principal value. The calculator gives one answer; the unit circle gives the second.
Wrong sign in the symmetry. Use CAST: only the relevant ratio is positive in each quadrant.
Mixing degree and radian mode. Sin(0.5) in radian mode ≠ sin(0.5°) in degree mode.
Forgetting to extend to the full interval. If the interval is bigger than 360°, add multiples of 360° (or 2π) to your two base solutions.
Discarding negative roots without checking the quadrant. cos α = ±4/5 — the sign depends on which quadrant α lies in.

Next note: Exact Values. The unit circle plus two special right triangles (30–60–90 and 45–45–90) give you exact values of sin, cos, and tan for every multiple of 30° and 45° — meaning answers in surds rather than decimals.

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