Mathematics Advanced • Year 11 • Module 2 • Lesson 9
Domains and Ranges of Trigonometric Functions
Apply domain and range analysis to AC circuit voltage bounds, tide-height modelling, light-intensity functions, range existence problems, and asymptote detection.
Problem 1 — AC voltage envelope (engineering)
An AC voltage in a domestic circuit is modelled by $V(t) = 240 \sin(100 \pi t) + 5$, where $V$ is in volts and $t$ is in seconds. (The "+5" represents a small DC offset.) Engineers care about the maximum and minimum voltage seen by appliances.
Set up: What are we solving for?
(i) Starting from $-1 \leq \sin(100\pi t) \leq 1$, derive the maximum and minimum values of $V(t)$. State the range of $V$. 3 marks
(ii) Is the function $V(t)$ defined for every real $t$? Explain in one sentence with reference to the domain of sine. 2 marks
(iii) The lesson notes that sine is bounded "because every appliance sees a known maximum stress". State a numerical voltage that the appliance can be guaranteed never to exceed. 1 mark
Stuck? Revisit lesson § Why this matters for physics callout.Problem 2 — Tide-height model (oceanographic)
A tide-height function is $H(t) = 4 - 3 \cos\left(\frac{\pi t}{6}\right)$, where $H$ is in metres and $t$ is in hours from midnight.
Set up: What are we solving for?
(i) Find the range of $H(t)$ over all real $t$, showing the inequality manipulation step-by-step. 3 marks
(ii) The local marina warns of tides below 2 m (boats sit on the seabed). Is this warning ever triggered by this model? Justify. 2 marks
(iii) A junior modeller writes "the range is $[1, 7]$ because we apply $\pm 3$ to the midline 4". Verify this matches your answer in (i) and explain in one sentence why the coefficient of $\cos$ being negative doesn't affect the range. 2 marks
Problem 3 — Light-intensity with asymptote (optical)
A laboratory laser's intensity at angle $\theta$ relative to a fixed axis is modelled as $I(\theta) = I_0 \sec^2 \theta$ (units of $I_0$). The detector saturates at $4 I_0$.
Set up: What are we solving for?
(i) State the domain of $I(\theta)$ in general-solution form. 2 marks
(ii) Find the smallest positive $\theta$ at which $I(\theta) = 4 I_0$ (i.e. $\sec^2 \theta = 4$). Show every step. 3 marks
(iii) Explain in 1-2 sentences why $I(\theta)$ is unbounded above (cite the range of $\sec$ and the lesson's bounded/unbounded distinction). 2 marks
Stuck on (ii)? $\sec^2 \theta = 4 \Rightarrow \cos \theta = \pm \frac{1}{2}$; smallest positive is $\theta = \frac{\pi}{3}$.Problem 4 — Does a solution exist?
For each equation, decide whether a real $\theta$ exists that satisfies it. Tick yes/no with one sentence of reasoning referencing the relevant range.
Set up: What are we solving for?
(i) $\sin \theta = \frac{3}{2}$ □ yes □ no Reason: 1 mark
(ii) $\sec \theta = 0.8$ □ yes □ no Reason: 1 mark
(iii) $\tan \theta = 10^{12}$ □ yes □ no Reason: 1 mark
(iv) $\cos \theta = -1$ □ yes □ no Reason: 1 mark
(v) $\csc \theta = 0$ □ yes □ no Reason: 2 marks
(vi) $\cot \theta = -10$ □ yes □ no Reason: 1 mark
Problem 5 — Transformed function domain audit
A signal-processing engineer must find the domain of $f(\theta) = \frac{1 + \cos \theta}{\sin \theta(1 - \sin \theta)}$ before computing anything else.
Set up: What are we solving for?
(i) List every condition the denominator imposes on $\theta$. 2 marks
(ii) Solve each condition over $\mathbb{R}$ (use general solutions). 3 marks
(iii) Write the domain of $f$ as a single set-builder description, then list every excluded value in $[0, 2\pi]$. 2 marks
Stuck? Set $\sin \theta = 0$ AND $1 - \sin \theta = 0$ as separate conditions.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — AC voltage
Set up. We're scaling and shifting the range of sine to find the voltage envelope.
(i) $-1 \leq \sin(100\pi t) \leq 1 \Rightarrow -240 \leq 240 \sin(100\pi t) \leq 240 \Rightarrow -235 \leq V \leq 245$. Range = $\mathbf{[-235, 245]}$ V. Max = 245 V, min = $-235$ V.
(ii) Yes — sine is defined for every real input, so $V(t)$ is defined for every real $t$.
(iii) The appliance can be guaranteed never to exceed 245 V (the upper bound of the range).
Problem 2 — Tide height
Set up. We're applying inequality manipulation with a negative coefficient and checking whether a safety threshold is breached.
(i) $-1 \leq \cos\left(\frac{\pi t}{6}\right) \leq 1$. Multiply by $-3$ (reverses inequality): $-3 \leq -3 \cos(\cdot) \leq 3$. Add 4: $1 \leq H \leq 7$. Range = $\mathbf{[1, 7]}$ m.
(ii) Minimum tide is 1 m (well below 2 m), so the warning is triggered — whenever $H(t) \leq 2$ m, the boats are at risk. Algebraically, $4 - 3\cos\left(\frac{\pi t}{6}\right) \leq 2 \Rightarrow \cos\left(\frac{\pi t}{6}\right) \geq \frac{2}{3}$ — achievable.
(iii) Matches: midline = 4, amplitude = 3, range = $[4 - 3, 4 + 3] = [1, 7]$. The negative coefficient on cos only reflects the wave shape; range is determined by $|a|$, not the sign of $a$.
Problem 3 — Light intensity
Set up. We're identifying domain restrictions, solving an intensity equation, and explaining unboundedness via the range of sec.
(i) $\sec \theta$ undefined when $\cos \theta = 0$, i.e. $\theta = \frac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$. Domain of $I(\theta)$: all real $\theta$ except $\mathbf{\theta = \frac{\pi}{2} + n\pi}$.
(ii) $\sec^2 \theta = 4 \Rightarrow \cos^2 \theta = \frac{1}{4} \Rightarrow \cos \theta = \pm \frac{1}{2}$. Smallest positive: $\cos \theta = \frac{1}{2}$ at $\theta = \frac{\pi}{3}$ (QI). So $\theta = \mathbf{\frac{\pi}{3}}$.
(iii) Range of $\sec \theta$ is $(-\infty, -1] \cup [1, \infty)$, so $\sec^2 \theta \in [1, \infty)$ — unbounded above. Hence $I(\theta) = I_0 \sec^2 \theta$ also has no upper bound: as $\theta$ approaches the asymptotic angles ($\frac{\pi}{2} + n\pi$), intensity grows without limit (consistent with the lesson's bounded/unbounded distinction).
Problem 4 — Solution existence
(i) No — $\sin \theta \in [-1, 1]$, so $\frac{3}{2}$ is outside.
(ii) No — $|\sec \theta| \geq 1$, so 0.8 is in the forbidden interval.
(iii) Yes — $\tan \theta$ has range all real $y$ (unbounded).
(iv) Yes — $\cos \theta = -1$ at $\theta = \pi + 2n\pi$.
(v) No — $\csc \theta = \frac{1}{\sin \theta}$ has range $(-\infty, -1] \cup [1, \infty)$; zero is not attainable (and the reciprocal of any non-zero number is non-zero). Alternative reasoning: $\csc \theta = 0 \Rightarrow \frac{1}{\sin \theta} = 0$ has no solution.
(vi) Yes — $\cot$ has range all real $y$ (unbounded).
Problem 5 — Domain audit
Set up. We're enumerating conditions from a denominator with two factors, then combining the exclusions.
(i) Need $\sin \theta \neq 0$ AND $1 - \sin \theta \neq 0$ (i.e. $\sin \theta \neq 1$).
(ii) $\sin \theta = 0 \Rightarrow \theta = n\pi$, $n \in \mathbb{Z}$. $\sin \theta = 1 \Rightarrow \theta = \frac{\pi}{2} + 2n\pi$, $n \in \mathbb{Z}$.
(iii) Domain: $\{\theta \in \mathbb{R} : \theta \neq n\pi \text{ and } \theta \neq \frac{\pi}{2} + 2n\pi, n \in \mathbb{Z}\}$. On $[0, 2\pi]$ excluded values: $\mathbf{\theta = 0, \frac{\pi}{2}, \pi, 2\pi}$ — four exclusions. ($\frac{3\pi}{2}$ is fine because $\sin \frac{3\pi}{2} = -1$, not 0 or 1.)