Mathematics Advanced • Year 11 • Module 2 • Lesson 3
The Unit Circle
Build procedural fluency in reading coordinates off the unit circle, applying ASTC, and using reference angles in all four quadrants.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the unit-circle identity: the point P on the unit circle at angle θ has coordinates
P(θ) = ( ____________ , ____________ ).
Q1.2 State which of sin, cos, tan are positive in each quadrant (use ASTC).
Quadrant I: ____________ Quadrant II: ____________
Quadrant III: ____________ Quadrant IV: ____________
Q1.3 Write the Pythagorean identity that comes from the equation of the unit circle:
____________ + ____________ = 1.
2. Worked example — coordinates at θ = 2π/3
Follow each line of algebra. Every step has a reason on the right.
Problem. Find the exact coordinates of the point on the unit circle corresponding to θ = 2π/3.
Step 1 — Identify the quadrant.
π/2 = 3π/6 < 4π/6 = 2π/3 < 6π/6 = π.
Reason: 2π/3 lies between π/2 and π, so it is in Quadrant II.
Step 2 — Find the reference angle α.
α = π − 2π/3 = 3π/3 − 2π/3 = π/3.
Reason: in QII, α = π − θ.
Step 3 — Use the exact values at the reference angle.
cos(π/3) = 1/2, sin(π/3) = √3/2.
Reason: from the 30-60-90 special triangle.
Step 4 — Apply ASTC signs.
In QII: only sine is positive ⇒ cosine negative, sine positive.
P(2π/3) = (−1/2, √3/2).
Conclusion. P(2π/3) = (−1/2, √3/2).
3. Faded example — using the Pythagorean identity
If sin θ = −3/5 and θ lies in Quadrant IV, find cos θ. Fill in each blank. 4 marks
Step 1 — Write the identity.
cos²θ + sin²θ = ________
Step 2 — Substitute sin θ and isolate cos²θ.
cos²θ + (____)² = 1 ⇒ cos²θ = 1 − ________ = ________
Step 3 — Take the square root (two cases).
cos θ = ± ________
Step 4 — Apply ASTC to choose the sign.
In QIV, cosine is ________________ (positive / negative). So cos θ = ________.
Conclusion. cos θ = ________________.
4. Graduated practice — coordinates, signs and reference angles
For each angle, find the exact unit-circle coordinates (cos θ, sin θ). State the quadrant and the reference angle.
Foundation — quadrantal angles (4 questions)
| Q | Angle θ | P(θ) = (cos θ, sin θ) | Quadrant or axis |
|---|---|---|---|
| 4.1 1 | 0 | ||
| 4.2 1 | π/2 | ||
| 4.3 1 | π | ||
| 4.4 1 | 3π/2 |
Standard — typical HSC difficulty (6 questions)
Show the reference angle and the ASTC sign reasoning for each.
4.5 Find P(5π/4) in exact form. State quadrant and reference angle. 2 marks
4.6 Find P(5π/6) in exact form. State quadrant and reference angle. 2 marks
4.7 Find P(−π/2) in exact form (use a coterminal angle if helpful). 2 marks
4.8 State the exact value of tan(4π/3). 2 marks
4.9 If cos θ = −5/13 and θ is in Quadrant II, find sin θ (exact form). 2 marks
4.10 If tan θ = 7/24 and θ is in Quadrant III, find sin θ and cos θ (exact form). 2 marks
Extension — combine concepts (2 questions)
4.11 A point P on the unit circle has coordinates (−√3/2, −1/2). Find the angle θ in [0, 2π) (exact form, in radians), and justify using ASTC and a reference angle. 3 marks
4.12 Show that for any angle θ, the point P(θ) on the unit circle satisfies x² + y² = 1. Then explain in one sentence why this is equivalent to the Pythagorean identity sin²θ + cos²θ = 1. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Unit-circle identity
P(θ) = (cos θ, sin θ). The x-coordinate is cosine, y-coordinate is sine.
Q1.2 — ASTC signs
QI: all (sin, cos, tan all positive). QII: sin only. QIII: tan only. QIV: cos only.
Q1.3 — Pythagorean identity
cos²θ + sin²θ = 1. (Equivalently sin²θ + cos²θ = 1; order doesn't matter.)
Q3 — Faded example: sin θ = −3/5, θ in QIV
Step 1: cos²θ + sin²θ = 1.
Step 2: cos²θ + (−3/5)² = 1 ⇒ cos²θ = 1 − 9/25 = 16/25.
Step 3: cos θ = ± 4/5.
Step 4: In QIV, cosine is positive. So cos θ = 4/5.
Conclusion: cos θ = 4/5.
Q4.1 — θ = 0
P(0) = (1, 0). Lies on the positive x-axis (boundary, not in any quadrant).
Q4.2 — θ = π/2
P(π/2) = (0, 1). Lies on the positive y-axis.
Q4.3 — θ = π
P(π) = (−1, 0). Lies on the negative x-axis.
Q4.4 — θ = 3π/2
P(3π/2) = (0, −1). Lies on the negative y-axis.
Q4.5 — θ = 5π/4
Quadrant III (since π < 5π/4 < 3π/2). Reference angle α = 5π/4 − π = π/4. From the 45-45-90 triangle: cos(π/4) = sin(π/4) = √2/2. In QIII both are negative. P(5π/4) = (−√2/2, −√2/2).
Q4.6 — θ = 5π/6
Quadrant II. Reference angle α = π − 5π/6 = π/6. cos(π/6) = √3/2, sin(π/6) = 1/2. In QII: cos negative, sin positive. P(5π/6) = (−√3/2, 1/2).
Q4.7 — θ = −π/2
Coterminal with −π/2 + 2π = 3π/2 (on negative y-axis). P(−π/2) = (0, −1).
Q4.8 — tan(4π/3)
Quadrant III (since π < 4π/3 < 3π/2). Reference angle α = 4π/3 − π = π/3. tan(π/3) = √3. In QIII, tangent is positive. tan(4π/3) = √3.
Q4.9 — cos θ = −5/13, θ in QII
sin²θ = 1 − 25/169 = 144/169, so sin θ = ± 12/13. In QII, sin is positive: sin θ = 12/13.
Q4.10 — tan θ = 7/24, θ in QIII
Build a 7-24-25 right triangle (since 7² + 24² = 49 + 576 = 625 = 25²). At the reference angle: sin = 7/25, cos = 24/25. In QIII both are negative: sin θ = −7/25, cos θ = −24/25. (Check: tan = (−7/25)/(−24/25) = 7/24 ✓.)
Q4.11 — Recover θ from coordinates
Both coordinates negative ⇒ QIII. |x| = √3/2 = cos(π/6); |y| = 1/2 = sin(π/6). So the reference angle is π/6. In QIII, θ = π + α = π + π/6 = 7π/6.
Q4.12 — Pythagorean identity from unit circle
The unit circle has equation x² + y² = 1 (radius 1 centred at origin). Every point P(θ) on it satisfies this equation. Since P(θ) = (cos θ, sin θ) by definition, substituting x = cos θ and y = sin θ gives cos²θ + sin²θ = 1 for every θ. The two equations describe the same geometric constraint — one in Cartesian coordinates, the other in trigonometric terms.