Mathematics Advanced • Year 11 • Module 2 • Lesson 3
The Unit Circle
Apply the unit-circle definition to GPS bearings, Ferris-wheel kinematics, lighthouse light tracking, and identity-based deduction problems.
Problem 1 — GPS satellite position (the lesson's callout)
A GPS satellite orbits the Earth in a near-circular path. To simplify, treat the orbit as a unit circle centred at the Earth and measure the satellite's angular position from the positive x-axis (zero longitude). Three checkpoints record the satellite at θ = π/6, θ = 7π/4, and θ = 2π/3 respectively.
Set up: What are we solving for?
(i) Find the exact unit-circle coordinates (cos θ, sin θ) for each of the three checkpoints. 3 marks
(ii) Without computing again, state cos(−π/6) and sin(−π/6). Briefly justify using a symmetry of the unit circle. 2 marks
(iii) The GPS receiver detects a position at (−1/2, √3/2). Which checkpoint — or what angle — corresponds to this reading? Justify using ASTC. 2 marks
Stuck on (ii)? cos is an even function (symmetric in x-axis); sin is an odd function.Problem 2 — Ferris wheel kinematics (modelling)
A small Ferris wheel of radius 8 m rotates anticlockwise. A passenger boards at the rightmost point (the 3-o'clock position). Let θ be the angle (in radians) the carriage has swept anticlockwise from the boarding position. The passenger's position is (8 cos θ, 8 sin θ) relative to the wheel's centre.
Set up: What are we solving for?
(i) Find the passenger's exact position (in metres) after the wheel has rotated by 5π/6 rad. State the quadrant. 2 marks
(ii) The wheel's centre is 10 m above the ground. Find the passenger's height above ground after rotating by 7π/6 rad (exact form). 2 marks
(iii) A second passenger sits diametrically opposite the first. Express the second passenger's angular position in two different ways: (a) using "θ + π" applied to the first passenger's angle, and (b) using the negation symmetry P(θ + π) = (−cos θ, −sin θ). Verify both descriptions agree when θ = 5π/6. 2 marks
Problem 3 — Lighthouse beam direction (data)
A lighthouse beam rotates at constant angular speed. At sample times t = 1, 2, 3, 4 seconds, an observer records the beam pointing at the unit-circle angles below. The observer asks you to compute the (x, y) direction vector of the beam at each time.
| t (s) | θ (rad) | (cos θ, sin θ) | Quadrant / axis |
|---|---|---|---|
| 1 | π/4 | ||
| 2 | 3π/4 | ||
| 3 | 5π/4 | ||
| 4 | 7π/4 |
Set up: What are we solving for?
(i) Fill in the (cos θ, sin θ) column with exact values. 4 marks
(ii) What is the angular speed of the beam in rad/s? Justify by examining the Δθ/Δt pattern across the four samples. 2 marks
(iii) Predict the (x, y) direction at t = 8 seconds. (Hint: find θ(8), reduce by 2π if necessary, then read off the value.) 2 marks
Stuck on (iii)? If Δθ per second is constant, θ(t) = θ(0) + ω t — back-extrapolate from t = 1.Problem 4 — Identity deduction (algebraic application)
The Pythagorean identity gives a free extra equation that lets you find missing trig ratios from one known value plus a quadrant.
Set up: What are we solving for?
(i) Given sin θ = 2/3 and θ in QII, find the exact values of cos θ and tan θ. 3 marks
(ii) Given cos θ = −4/5 and tan θ > 0, identify the quadrant of θ, then find sin θ. 3 marks
(iii) Without using a calculator, verify the identity cos²θ + sin²θ = 1 explicitly for θ = π/3 using the special-triangle exact values. 2 marks
Problem 5 — Quadrant claim evaluation
A student claims: "sin θ is always positive when θ is between 0 and π, and always negative when θ is between π and 2π." Evaluate this claim with explicit examples.
Set up: What are we solving for?
(i) Test the first half of the claim by computing sin(π/6), sin(π/2), sin(5π/6) using exact values. 2 marks
(ii) Test the second half by computing sin(7π/6), sin(3π/2), sin(11π/6). 2 marks
(iii) Is the claim (a) true everywhere in the stated intervals, (b) false everywhere, or (c) almost-but-not-quite true? Identify any boundary or special cases (think about θ = 0, π, 2π) and rewrite the claim more carefully. 2 marks
Stuck on (iii)? At the boundaries (θ = 0, π, 2π), sin θ = 0 — it's neither positive nor negative.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — GPS satellite
Set up. We read coordinates off the unit circle for given angles using reference angles + ASTC, then exploit even/odd symmetry to invert without recomputation.
(i) θ = π/6 (QI): cos = √3/2, sin = 1/2. P = (√3/2, 1/2).
θ = 7π/4 (QIV): ref π/4; cos pos, sin neg. P = (√2/2, −√2/2).
θ = 2π/3 (QII): ref π/3; cos neg, sin pos. P = (−1/2, √3/2).
(ii) cos is even: cos(−π/6) = cos(π/6) = √3/2. sin is odd: sin(−π/6) = −sin(π/6) = −1/2. Justification: P(−θ) reflects P(θ) in the x-axis, leaving x-coordinate unchanged and negating the y-coordinate.
(iii) The reading (−1/2, √3/2) matches checkpoint θ = 2π/3 from (i). The x-coordinate is negative and y positive, so we're in QII; |x| = 1/2 and |y| = √3/2 correspond to reference angle π/3; so θ = π − π/3 = 2π/3. ✓
Problem 2 — Ferris wheel
Set up. The passenger's position on a circle of radius 8 is (8 cos θ, 8 sin θ); we apply ASTC and the special-triangle values, then add the centre's height to convert to ground level.
(i) θ = 5π/6 (QII, ref π/6): cos = −√3/2, sin = 1/2. Position = (8 × −√3/2, 8 × 1/2) = (−4√3, 4) m. Quadrant II.
(ii) θ = 7π/6 (QIII, ref π/6): sin = −1/2. y-coordinate (relative to wheel centre) = 8 × −1/2 = −4 m. Height above ground = 10 + (−4) = 6 m.
(iii) (a) Second passenger's angle = first's angle + π. At θ = 5π/6, second is at 5π/6 + π = 11π/6. P(11π/6) (QIV, ref π/6) = (√3/2, −1/2), scaled by 8 gives (4√3, −4) m.
(b) Using symmetry: P(θ + π) = (−cos θ, −sin θ) = (−(−√3/2), −(1/2)) = (√3/2, −1/2), scaled by 8 gives (4√3, −4) m. ✓ Both methods agree.
Problem 3 — Lighthouse beam direction
Set up. The angles increase by π/2 per second; we read coordinates at each sample, then extrapolate forward in time using the linear pattern.
(i) t = 1, θ = π/4 (QI): P = (√2/2, √2/2).
t = 2, θ = 3π/4 (QII): P = (−√2/2, √2/2).
t = 3, θ = 5π/4 (QIII): P = (−√2/2, −√2/2).
t = 4, θ = 7π/4 (QIV): P = (√2/2, −√2/2).
(ii) Δθ per second = (3π/4 − π/4) = π/2 rad/s. Same gap holds between every adjacent pair. So ω = π/2 rad/s.
(iii) θ(8) = θ(1) + (8 − 1) × π/2 = π/4 + 7π/2 = π/4 + 14π/4 = 15π/4. Reduce mod 2π = 8π/4: 15π/4 − 8π/4 = 7π/4. So at t = 8, θ = 7π/4 (same as t = 4). Direction = (√2/2, −√2/2).
Problem 4 — Identity deduction
Set up. Use cos²θ + sin²θ = 1 to find a missing magnitude, then ASTC to pin down the sign.
(i) cos²θ = 1 − 4/9 = 5/9, so cos θ = ±√5/3. QII: cos negative ⇒ cos θ = −√5/3. Then tan θ = sin/cos = (2/3) / (−√5/3) = −2/√5 = −2√5/5 (rationalised).
(ii) cos < 0 ⇒ QII or QIII. tan > 0 ⇒ QI or QIII. Intersection: QIII. sin²θ = 1 − 16/25 = 9/25, so sin θ = ±3/5; in QIII, sin negative ⇒ sin θ = −3/5. (Check: tan = (−3/5)/(−4/5) = 3/4 > 0 ✓.)
(iii) cos(π/3) = 1/2, sin(π/3) = √3/2. cos²(π/3) + sin²(π/3) = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1 ✓. The identity holds.
Problem 5 — Claim evaluation
Set up. We test the claim at sample angles in each interval, then check what happens at the boundaries.
(i) sin(π/6) = 1/2 (+). sin(π/2) = 1 (+). sin(5π/6) = sin(π − 5π/6) = sin(π/6) = 1/2 (+). All positive — consistent with the first half.
(ii) sin(7π/6) (QIII, ref π/6) = −1/2 (−). sin(3π/2) = −1 (−). sin(11π/6) (QIV, ref π/6) = −1/2 (−). All negative — consistent with the second half.
(iii) The claim is almost-but-not-quite true. The interior of each interval is correctly described, but the endpoints θ = 0, π, 2π (and any equivalent) give sin θ = 0, which is neither positive nor negative. A correctly stated version: "sin θ > 0 for θ in (0, π), sin θ < 0 for θ in (π, 2π), and sin θ = 0 at θ = 0, π, 2π." — using open intervals.