Mathematics Advanced • Year 11 • Module 2 • Lesson 15
Review and Connections
Practise HSC-style writing on three pillars at once: exact values + identity, transformed graph sketching with range, and a multi-step modelling-and-solving extended response.
1. Short-answer questions
1.1 If $\sin\theta = 1/3$ and $\pi/2 < \theta < \pi$, find the exact value of $\tan\theta$. Justify the sign using the quadrant. 3 marks Band 3-4
1.2 Sketch $y = 2\sin\!\left(x + \dfrac{\pi}{3}\right) - 1$ for $-\dfrac{\pi}{3} \leq x \leq \dfrac{5\pi}{3}$. State the range and label the maximum, minimum, midline crossings, and the $x$-intercepts within that domain. 4 marks Band 4
1.3 How many solutions does $2\cos x = 1$ have in $0 \leq x \leq 4\pi$? Justify with a period count, then list all the exact solutions. 3 marks Band 4
Stuck on 1.3? Rewrite as $\cos x = 1/2$; period of cosine is $2\pi$, so $[0, 4\pi]$ contains 2 periods.2. Extended response — full-module modelling
2.1 A coastal town's tide height (in metres) is modelled by $h(t) = 3 + 1.5\cos\!\left(\dfrac{2\pi t}{12}\right)$ where $t$ is hours since the morning high tide.
(a) State the high and low tide values, the period, and the midline. Identify the geometric meaning of each: which physical features of the tide cycle do they correspond to?
(b) A small fishing boat needs $h \geq 3.75$ m to leave the harbour safely. Find the first time $t > 0$ when $h(t) = 3.75$, and use the symmetry of cosine about $t = 0$ to determine the total time per 12-hour cycle during which the boat can leave.
(c) A student writes: "Since the high tide is at $t = 0$ and again at $t = 12$, the boat can leave every 12 hours — once per day." Evaluate this claim. Compute how many complete "boat-leaving windows" there are in a 24-hour day, and identify which of the lesson's three Traps the student has fallen into. Then list one real-world limitation of the simple cosine tide model. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — high $= 4.5$ m, low $= 1.5$ m, period $= 12$ h, midline $= 3$ m.
• 1 mark — geometric meanings: high $=$ peak of cosine; low $=$ trough; period $= $ time between consecutive high tides; midline $= $ average tide height (the "sea-level baseline").
Part (b) — 3 marks
• 1 mark — sets up $3 + 1.5\cos(\pi t/6) = 3.75 \Rightarrow \cos(\pi t/6) = 1/2$.
• 1 mark — first positive: $\pi t/6 = \pi/3 \Rightarrow t = 2$ h.
• 1 mark — total window per cycle: by cosine symmetry about $t = 0$, the boat can leave for $t \in [-2, 2]$ relative to a high tide, total $4$ hours per 12-hour cycle.
Part (c) — 3 marks
• 1 mark — identifies that there are actually 2 high tides per 24 hours (period is 12 h), so 2 windows per day.
• 1 mark — names Trap 02 from L13 (Forgetting to account for multiple cycles) — alternatively the daylight/period-counting trap.
• 1 mark — limitation: real tides have a 12 h 25 min period (semidiurnal), spring/neap modulation, weather effects, or do not always have equal high tides.
Your response:
Stuck on (c)? Period 12 h means 2 cycles fit in 24 h. Each cycle has 1 leaving window. So 2 windows per day.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\sin\theta = 1/3$ in Q2 (3 marks)
Sample response. $\cos^2\theta = 1 - 1/9 = 8/9$ [1]. In Q2, cosine is negative, so $\cos\theta = -2\sqrt{2}/3$ [1]. $\tan\theta = \sin\theta/\cos\theta = (1/3)/(-2\sqrt{2}/3) = -1/(2\sqrt{2}) = $ $-\sqrt{2}/4$ [1].
Marking notes. 1 mark Pythagorean identity. 1 mark for the sign — explicit quadrant reference required. 1 mark for the final exact value (rationalised denominator preferred but not required).
1.2 — Sketch $y = 2\sin(x + \pi/3) - 1$ (4 marks)
Sample response. Rewrite as $y = 2\sin(x - (-\pi/3)) - 1$: amplitude $2$, period $2\pi$, phase shift $\pi/3$ left, vertical shift $1$ down. Range $= [-1 - 2, -1 + 2] = $ $[-3, 1]$ [1]. Key points within $[-\pi/3, 5\pi/3]$:
At $x = -\pi/3$ (phase-shift point): midline going up, $y = -1$ [0.5].
Max at $x = -\pi/3 + \pi/2 = \pi/6$: $y = 1$ [0.5].
Midline crossing at $x = -\pi/3 + \pi = 2\pi/3$: $y = -1$ [0.5].
$x$-intercepts where $2\sin(x + \pi/3) - 1 = 0 \Rightarrow \sin(x + \pi/3) = 1/2$. In Q1: $x + \pi/3 = \pi/6 \Rightarrow x = -\pi/6$. In Q2: $x + \pi/3 = 5\pi/6 \Rightarrow x = \pi/2$ [1].
Min at $x = -\pi/3 + 3\pi/2 = 7\pi/6$: $y = -3$ [0.5].
Marking notes. 1 mark for range. 2 marks for max/min/midline labels. 1 mark for the two $x$-intercepts in the domain.
1.3 — $2\cos x = 1$ on $[0, 4\pi]$ (3 marks)
Sample response. $\cos x = 1/2$ [1]. Period of cosine is $2\pi$, so $[0, 4\pi]$ contains 2 periods; 2 solutions per period $\Rightarrow$ 4 solutions [1]. Exact list: in $[0, 2\pi]$, $x = \pi/3$ and $x = 5\pi/3$. Adding $2\pi$ to each: $x = 7\pi/3$ and $x = 11\pi/3$. $x = \pi/3, 5\pi/3, 7\pi/3, 11\pi/3$ [1].
Marking notes. 1 mark for rewriting. 1 mark for the period count. 1 mark for all four exact solutions (lose 0.5 for any missing one).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Parameters and physical meaning. High $= 3 + 1.5 = $ $4.5$ m, low $= 3 - 1.5 = $ $1.5$ m, period $= 2\pi / (2\pi/12) = $ $12$ h, midline $= $ $3$ m. [1 mark]
Physical meanings: high tide $4.5$ m is the peak depth above harbour datum; low tide $1.5$ m is the minimum depth; period $12$ h is the time between consecutive high tides (semidiurnal model); midline $3$ m is the mean tide level (the "sea-level baseline" for the harbour). [1 mark]
Part (b) — When the boat can leave. Set $3 + 1.5\cos(\pi t/6) = 3.75 \Rightarrow \cos(\pi t/6) = 0.75/1.5 = 1/2$. [1 mark]
First positive: $\pi t/6 = \pi/3 \Rightarrow t = $ $2$ h after high tide. [1 mark]
By the symmetry of cosine about $t = 0$, the tide is $\geq 3.75$ for $t \in [-2, 2]$ relative to each high tide — total $4$ hours per 12-hour cycle. [1 mark]
Part (c) — Evaluating the student's claim. The claim is incorrect. The model has period $12$ h, so a $24$-hour day contains two complete tide cycles, each with its own high tide and its own boat-leaving window. So there are 2 windows per day (one centred on the morning high tide, one centred on the afternoon high tide ~12 h later), not 1. [1 mark]
This is exactly Trap 02 from Lesson 13 — "Forgetting to account for multiple cycles." The student counted one cycle and stopped, instead of counting how many cycles fit in the given $24$-hour domain. [1 mark — Trap 02 named]
Limitations of the simple cosine tide model (any one): real tides have period $\approx 12$ h $25$ min (not exactly $12$), which gradually shifts high tides through the day; spring/neap variations modulate amplitude over a $\approx 14$-day cycle as the moon's gravity adds to or subtracts from the sun's; and weather (storm surges, atmospheric pressure changes) can shift levels by up to a metre, none of which appears in the model. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes parameters in (a) without physical meaning; finds $t = 2$ in (b) without considering symmetry; doesn't dispute the student's claim. ≈ 3-4 marks.
Band 4: Full (a) with meanings; (b) including the symmetry argument; (c) identifies "2 windows per day" but doesn't name a Trap. ≈ 5-6 marks.
Band 5: All of the above with explicit Trap 02 reference and one specific real-world limitation. ≈ 7 marks.
Band 6: All of the above plus a precise quantitative limitation (e.g. naming the $12$ h $25$ min lunar tidal period, or the $14$-day spring/neap cycle). 8/8.