Key Idea: The unit circle turns sin, cos and tan into coordinates, so they work for any angle — not just acute ones. It powers the exact special-angle values and the ambiguous case, both pure Paper 1, non-calculator.
⭕ The unit circle: (cos θ, sin θ)
- the x-coordinate of the point
- the y-coordinate of the point
- angle measured anticlockwise from the positive x-axis
On a circle of radius 1 the point at angle θ is (cos θ, sin θ) — cos is the x-coordinate, sin is the y-coordinate. That single fact gives you signs by quadrant and the related-angle rules below.
📐 Exact values to know cold (Paper 1)
Tip: sin and cos swap across 45°: sin 30° = cos 60° = ½, and sin 60° = cos 30° = √3⁄2. Learn one column and mirror it. Also tan = sin ÷ cos, so you can rebuild the tan column.
🧭 Signs by quadrant (CAST)
Two-step method for any angle: find the acute reference angle, read its exact value, then attach the sign CAST gives for that quadrant. e.g. cos 150° is in Q2 (only sin positive) → negative.
🔁 Related (supplementary) angles
Tip: Because sin(180° − θ) = sin θ, two different angles share the same sine. That is exactly what makes the ambiguous case (finding an angle with the sine rule) produce two triangles.
✏️ IB-style worked examples
IB-style question — exact value via reference angle and quadrant
Find the exact value of cos(5π/6) without a calculator.
Step by step:
5π/6 = 150°, which lies in the second quadrant. Find the reference angle to the x-axis.
The reference value is cos 30°.
In Q2 only sine is positive (CAST), so cosine is negative.
cos(5π/6) = −√3⁄2
IB-style question — find sin θ from cos θ (acute)
Given cos θ = 1/4 with θ acute, find the exact value of sin θ.
Step by step:
Use the Pythagorean identity.
θ is acute, so sin θ is positive — take the positive root.
sin θ = √15⁄4
IB-style question — the ambiguous case (two triangles)
In triangle ABC, a = 8, A = 35° and b = 11. Find both possible values of angle B, and decide whether each is valid.
Step by step:
Apply the sine rule to find sin B.
Two angles share this sine: the acute value and its supplement.
Check the angle sum for each. With A = 35°: 35° + 127.9° = 162.9° < 180°, so both leave a positive third angle.
B ≈ 52.1° or B ≈ 127.9° — both triangles are valid.
Important: Two traps. (1) Quoting the exact value but dropping the CAST sign — cos 150° is −√3⁄2, not +√3⁄2. (2) When the sine rule gives an angle, your calculator hands you only the acute one — always test the supplement 180° − θ too.
Tap each card to reveal the answer.
Exam Tips
- Memorise the exact-value table for 0/30/45/60/90° in both degrees and radians — Paper 1 expects it instantly.
- Reference angle gives the value; CAST gives the sign. Do both, in that order.
- A → Q1, S → Q2, T → Q3, C → Q4 for which ratio is positive.
- sin(180° − θ) = sin θ: every sine has a supplementary partner — the heart of the ambiguous case.
- Finding an angle with the sine rule? Check 180° − θ, then keep it only if (known angle) + (obtuse) < 180°.