Mathematics Advanced • Year 11 • Module 2 • Lesson 3
The Unit Circle
Practise HSC-style writing on unit-circle coordinates, missing trig ratios, and a derivation of the Pythagorean identity from the equation of the unit circle.
1. Short-answer questions
1.1 (a) State the exact coordinates of the point on the unit circle corresponding to θ = 3π/4. (b) Hence write down the exact values of sin(3π/4) and cos(3π/4). 3 marks Band 3
1.2 If cos θ = −5/13 and θ is in Quadrant II, find the exact value of sin θ. Show all working using the Pythagorean identity. 3 marks Band 4
1.3 Without using a calculator, find the exact value of:
(a) sin(π + π/6) using the QIII symmetry P(π + α) = (−cos α, −sin α).
(b) cos(2π − π/4) using the QIV symmetry P(2π − α) = (cos α, −sin α). 4 marks Band 4-5
2. Extended response
2.1 Let P(θ) denote the point on the unit circle obtained by rotating the point (1, 0) by an angle θ (measured anticlockwise from the positive x-axis).
(a) Starting from the definition of the unit circle (x² + y² = 1) and the definitions cos θ = x-coordinate of P(θ) and sin θ = y-coordinate of P(θ), derive the Pythagorean identity sin²θ + cos²θ = 1. Justify that the identity holds for every angle θ, not just those in [0, 2π).
(b) Using only the unit-circle definition and ASTC (no calculator), find the exact value of cos(7π/6) and sin(7π/6). Show your reasoning explicitly.
(c) Given that tan θ = −5/12, determine the two possible values of cos θ in exact form, and identify the quadrant of θ corresponding to each value. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — identifies P(θ) = (cos θ, sin θ) and notes P(θ) lies on the unit circle, so (cos θ)² + (sin θ)² = 1.
• 1 mark — rewrites as sin²θ + cos²θ = 1 in standard notation.
• 1 mark — argues that for any θ (positive, negative, large), the point P(θ) is by definition always on the unit circle (coterminal angles land at the same point), so the identity holds universally.
Part (b) — 2 marks
• 1 mark — identifies 7π/6 in QIII with reference angle π/6 (since 7π/6 − π = π/6).
• 1 mark — correctly applies ASTC (both cos and sin negative in QIII): cos(7π/6) = −√3/2, sin(7π/6) = −1/2.
Part (c) — 3 marks
• 1 mark — identifies that tan θ < 0 means θ lies in QII or QIV (two quadrants where sin and cos have opposite signs).
• 1 mark — using a 5-12-13 right triangle (or the identity 1 + tan²θ = sec²θ), finds |cos θ| = 12/13.
• 1 mark — gives both values with quadrants: cos θ = −12/13 in QII, cos θ = 12/13 in QIV.
Your response:
Stuck on (c)? Recognise 5, 12, 13 as a Pythagorean triple — the hypotenuse is 13.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Coordinates at θ = 3π/4 (3 marks)
Sample response. 3π/4 lies in QII with reference angle α = π − 3π/4 = π/4. From the 45-45-90 triangle, cos(π/4) = sin(π/4) = √2/2. In QII, cos is negative and sin is positive.
(a) P(3π/4) = (−√2/2, √2/2).
(b) Therefore cos(3π/4) = −√2/2 and sin(3π/4) = √2/2.
Marking notes. 1 mark — correct coordinates (a). 1 mark — both values stated in (b) with rationalised form. 1 mark — explicit reference angle / ASTC justification (not just "from the unit circle"). Common error: writing (√2/2, −√2/2) by confusing x with y (Trap 01).
1.2 — sin θ from cos θ = −5/13 in QII (3 marks)
Sample response. By the Pythagorean identity, sin²θ + cos²θ = 1:
sin²θ = 1 − (−5/13)² = 1 − 25/169 = 144/169.
So sin θ = ±12/13. In QII, sine is positive, so sin θ = 12/13.
Marking notes. 1 mark — correct substitution into the identity. 1 mark — correct algebra to sin²θ = 144/169. 1 mark — takes the positive root with explicit ASTC justification ("in QII, sine is positive"). Common error: leaves answer as ±12/13 without choosing the sign (loses the final mark).
1.3 — Symmetry applications (4 marks)
Sample response.
(a) sin(π + π/6) using P(π + α) = (−cos α, −sin α): sin part = −sin(π/6) = −1/2. (Confirm: π + π/6 = 7π/6, which is in QIII where sin is negative; |sin| = 1/2 from the reference angle. ✓)
(b) cos(2π − π/4) using P(2π − α) = (cos α, −sin α): cos part = cos(π/4) = √2/2. (Confirm: 2π − π/4 = 7π/4, which is in QIV where cos is positive; |cos| = √2/2 from the reference angle. ✓)
Marking notes. 2 marks each part: 1 for correctly identifying which coordinate of the transformed P to read; 1 for the exact value with sign. Award 0.5 if a student finds the right value via reference angle + ASTC without using the supplied symmetry rule.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Derivation of the Pythagorean identity. By definition, P(θ) = (cos θ, sin θ) is the point reached by rotating (1, 0) anticlockwise by an angle θ on the unit circle. The unit circle has equation x² + y² = 1, so every point on it satisfies this equation. Substituting x = cos θ, y = sin θ:
(cos θ)² + (sin θ)² = 1, i.e. cos²θ + sin²θ = 1. [1 mark + 1 mark]
This holds for any θ: for negative θ, the rotation simply runs clockwise and lands on a point still on the unit circle. For θ outside [0, 2π), the rotation winds around one or more times but lands on a coterminal point, which is again on the unit circle. The identity is therefore universal. [1 mark]
Part (b) — cos(7π/6), sin(7π/6). Since π < 7π/6 < 3π/2, the angle lies in Quadrant III. Reference angle: α = 7π/6 − π = π/6. [1 mark]
From the 30-60-90 special triangle: cos(π/6) = √3/2, sin(π/6) = 1/2. In QIII, both cos and sin are negative (only tan positive):
cos(7π/6) = −√3/2, sin(7π/6) = −1/2. [1 mark]
Part (c) — Two possible values of cos θ given tan θ = −5/12. Because tan θ < 0, sin θ and cos θ have opposite signs. The ASTC quadrants where this happens are QII (sin positive, cos negative) and QIV (sin negative, cos positive). [1 mark]
Build a right triangle with opposite = 5 and adjacent = 12 (matching the magnitude of tan); by Pythagoras, the hypotenuse is √(25 + 144) = √169 = 13. So in absolute value, sin θ = 5/13 and cos θ = 12/13, giving |cos θ| = 12/13. [1 mark]
Combining with the quadrant analysis:
• If θ in QII: cos θ = −12/13.
• If θ in QIV: cos θ = 12/13. [1 mark] ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Quotes the identity in (a) without derivation; computes (b) using a calculator approximation rather than exact values; in (c) gives only one value of cos θ without acknowledging the two-quadrant ambiguity. ≈ 3-4 marks.
Band 4: Derives (a) by substitution but does not address the universality claim; (b) correct; (c) finds both values but does not label which quadrant each belongs to. ≈ 5-6 marks.
Band 5: All parts correct; quadrant labelling clear in (c); the universality argument in (a) given but brief. ≈ 7 marks.
Band 6: Full derivation including the universality argument; (b) explicitly invokes reference angle and ASTC (not just "from the unit circle"); (c) clearly distinguishes the two cases with reasoning. 8/8. Top scripts use the Pythagorean triple (5, 12, 13) confidently and mention 1 + tan²θ = sec²θ as an alternative path.