Mathematics Advanced • Year 11 • Module 2 • Lesson 10
Graphs of Sine and Cosine
Practise HSC-style writing on sinusoidal sketching, parameter recovery from features, and a structured derivation of the range formula from $y = a \sin(bx) + d$.
1. Short-answer questions
1.1 For $y = 3 \cos(2x)$, state (a) the amplitude, (b) the period (in radians), and (c) the range. 3 marks Band 3
1.2 Sketch one full cycle of $y = -2 \sin x$ on $0 \leq x \leq 2\pi$. Label the five key points $(x, y)$ on your sketch and state the amplitude and range. 4 marks Band 3-4
1.3 The graph of $y = a \cos(bx) + d$ has a maximum of 5, a minimum of $-1$, and a period of $\frac{2\pi}{3}$. Find positive values of $a$, $b$, and $d$. Show all working. 3 marks Band 4
Stuck on 1.3? Midline $d = \frac{\max + \min}{2}$; amplitude $|a| = \frac{\max - \min}{2}$; $b = \frac{2\pi}{\text{period}}$.2. Extended response
2.1 An acoustic engineer is modelling the air-pressure fluctuation near a speaker using a sinusoid $p(t) = a \sin(b t) + d$.
(a) Starting from the range of $\sin(b t)$, derive a general formula for the range of $y = a \sin(b t) + d$ in terms of $a$ and $d$ (treat $a$ and $b$ as positive). Justify each inequality step.
(b) The engineer measures pressure values that oscillate between $40$ Pa (minimum) and $80$ Pa (maximum), with one full cycle every $\frac{1}{200}$ s. Find positive $a$, $b$, and $d$ for the model. Show every step.
(c) Sketch one cycle of $p(t)$ for $0 \leq t \leq \frac{1}{200}$ s, labelling the maximum, minimum, and midline crossings on labelled axes. Then, in 1-2 sentences, evaluate a junior engineer's claim that "doubling $b$ doubles the period". Reference Trap 1 from the lesson explicitly. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — states $-1 \leq \sin(bt) \leq 1$ for all $t$.
• 1 mark — multiplies by $a$ (positive) to get $-a \leq a \sin(bt) \leq a$.
• 1 mark — adds $d$ to obtain $d - a \leq p(t) \leq d + a$, i.e. range $[d - a, d + a]$ (general form: $[d - |a|, d + |a|]$).
Part (b) — 3 marks
• 1 mark — midline $d = \frac{80 + 40}{2} = 60$ Pa.
• 1 mark — amplitude $a = \frac{80 - 40}{2} = 20$ Pa.
• 1 mark — $b = \frac{2\pi}{1/200} = 400\pi$ rad/s.
Part (c) — 2 marks
• 1 mark — sketch shows: starts at midline (60 Pa) at $t = 0$, rises to max (80) at $t = \frac{1}{800}$, returns to midline at $\frac{1}{400}$, falls to min (40) at $\frac{3}{800}$, returns to midline at $\frac{1}{200}$. Labels on key points.
• 1 mark — refutes the claim: doubling $b$ halves the period (period $= \frac{2\pi}{b}$ is inversely proportional to $b$). Explicit reference to Trap 1 ("larger $b$ means shorter period").
Your response:
Stuck on (a)? Multiply through $-1 \leq \sin(bt) \leq 1$ by $a$ (positive), then add $d$.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Parameters of $3 \cos(2x)$ (3 marks)
Sample response. $a = 3$, $b = 2$, $d = 0$. (a) Amplitude $= |a| = \mathbf{3}$. (b) Period $= \frac{2\pi}{|b|} = \frac{2\pi}{2} = \mathbf{\pi}$. (c) Range $= [d - |a|, d + |a|] = \mathbf{[-3, 3]}$.
Marking notes. 1 mark each for amplitude, period, range. Common error: stating period as $2\pi$ (forgetting the $b = 2$ compresses the graph — Trap 1). Award 0 if amplitude given as $-3$ instead of $3$ (Trap 2).
1.2 — Sketch $y = -2 \sin x$ (4 marks)
Sample response. $a = -2$, $b = 1$, $d = 0$. Amplitude $= |-2| = 2$. Range $= [-2, 2]$. The negative sign reflects the sine graph in the $x$-axis. Key points: $(0, 0)$, $\left(\frac{\pi}{2}, -2\right)$, $(\pi, 0)$, $\left(\frac{3\pi}{2}, 2\right)$, $(2\pi, 0)$. Sketch: a smooth wave starting at origin, going down first (reflected) to $-2$ at $\frac{\pi}{2}$, back to $0$ at $\pi$, up to 2 at $\frac{3\pi}{2}$, back to $0$ at $2\pi$.
Marking notes. 1 mark for amplitude. 1 mark for range. 1 mark for the five labelled points (signs correct — the reflection is essential). 1 mark for a smooth sketch that visibly goes down first (reflecting the negative $a$). Common error: drawing the standard sine going up first — costs the sketch mark.
1.3 — Find $a$, $b$, $d$ (3 marks)
Sample response. Midline $d = \frac{\max + \min}{2} = \frac{5 + (-1)}{2} = \mathbf{2}$.
Amplitude $|a| = \frac{\max - \min}{2} = \frac{5 - (-1)}{2} = 3$, so $a = \mathbf{3}$ (positive).
$b = \frac{2\pi}{\text{period}} = \frac{2\pi}{2\pi/3} = \mathbf{3}$.
So $y = 3 \cos(3x) + 2$. Check: max $= 3 + 2 = 5$ ✓, min $= -3 + 2 = -1$ ✓, period $= \frac{2\pi}{3}$ ✓.
Marking notes. 1 mark each for $d$, $a$, $b$. Working must show the formula used; bald answers without working cap at 2/3. A verification check is good practice but not required for full marks.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Derivation of the range formula. For all real $t$, $-1 \leq \sin(b t) \leq 1$ (the range of sine is independent of $b$). [1 mark]
Multiply through by $a > 0$ (positive, so inequalities preserve direction): $-a \leq a \sin(bt) \leq a$. [1 mark]
Add $d$ to every part: $d - a \leq a \sin(bt) + d \leq d + a$. So range of $p(t) = \mathbf{[d - a, d + a]}$. (For general $a$ — possibly negative — replace $a$ with $|a|$; the negative $a$ only reflects the wave in the midline, not the range.) [1 mark]
Part (b) — Find $a$, $b$, $d$. Midline: $d = \frac{80 + 40}{2} = \mathbf{60}$ Pa. [1 mark]
Amplitude: $a = \frac{80 - 40}{2} = \mathbf{20}$ Pa (taking positive $a$). [1 mark]
Angular frequency: $b = \frac{2\pi}{\text{period}} = \frac{2\pi}{1/200} = \mathbf{400\pi}$ rad/s. [1 mark]
Equation: $p(t) = 20 \sin(400 \pi t) + 60$ Pa.
Part (c) — Sketch and student-claim critique. Sketch one cycle over $\left[0, \frac{1}{200}\right]$ s. Five key points: $(0, 60)$ (midline, going up), $\left(\frac{1}{800}, 80\right)$ (max), $\left(\frac{1}{400}, 60\right)$ (midline, going down), $\left(\frac{3}{800}, 40\right)$ (min), $\left(\frac{1}{200}, 60\right)$ (midline, completing cycle). Axes labelled $t$ (s) and $p$ (Pa); horizontal dashed line at midline 60. [1 mark]
The student's claim ("doubling $b$ doubles the period") is incorrect — it commits Trap 1 from the lesson. Because period $= \frac{2\pi}{b}$, doubling $b$ halves the period (e.g. $b = 400\pi$ gives period $\frac{1}{200}$ s; $b = 800\pi$ gives period $\frac{1}{400}$ s). Larger $b$ means shorter period. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: (a) states the formula without derivation; (b) gets $d = 60$ but errors on $a$ or $b$; (c) sketch missing labels and Trap 1 not addressed. ≈ 3-4 marks.
Band 4: (a) one inequality step shown; (b) all three correct but no working; (c) sketch present but unlabelled; critique vague. ≈ 5-6 marks.
Band 5: (a) full derivation; (b) full working; (c) clean sketch with labels; identifies Trap 1 without numerical counter-example. ≈ 7 marks.
Band 6: Full rigour throughout: explicit positivity argument in (a), all three parameters with formulas in (b), labelled sketch with five key points, and critique with a numerical counter-example of the doubling-period claim. 8/8.