Mathematics Advanced • Year 11 • Module 2 • Lesson 9
Domains and Ranges of Trigonometric Functions
Practise HSC-style writing on transformed-tangent domains, transformed-cosine ranges, and a derivation of the range of secant from the range of cosine.
1. Short-answer questions
1.1 (a) State the domain of $y = \tan 2x$ in general-solution form. (b) Find every value of $x$ in $[0, \pi]$ where $\tan 2x$ is undefined. 3 marks Band 3
1.2 Find the range of $y = 4 - 2 \cos x$. Show the inequality manipulation step-by-step. 3 marks Band 3-4
1.3 Find the domain and range of $f(x) = \frac{1}{1 + \cos x}$. Use general-solution form for the domain. 4 marks Band 4
Stuck on 1.3 range? $1 + \cos x \in [0, 2]$, but excluding 0; so $\frac{1}{1 + \cos x} \geq \frac{1}{2}$.2. Extended response
2.1 A physics teacher wants a rigorous, step-by-step explanation of why secant has the range it does — not just a memorised statement.
(a) Starting from the unit-circle definition of cosine, state the range of $\cos x$ and explain (in 1-2 sentences) why this range is exactly $[-1, 1]$.
(b) Derive the range of $\sec x = \frac{1}{\cos x}$. Use case analysis on $\cos x > 0$ and $\cos x < 0$ separately, and pay careful attention to which inequalities flip when taking reciprocals. State why $\cos x = 0$ must be excluded.
(c) Hence find the range of $y = 3 \sec x - 1$. Then, in 1-2 sentences, evaluate the following student claim: "the range of any function of the form $A \sec x + B$ is the whole real line, because secant has asymptotes". Reference Trap 2 from the lesson explicitly. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — states range $[-1, 1]$.
• 1 mark — justifies via unit circle: $\cos x$ is the $x$-coordinate of a point on the unit circle, which satisfies $-1 \leq x \leq 1$.
Part (b) — 4 marks
• 1 mark — excludes $\cos x = 0$ (would force division by zero), so works with $\cos x \in [-1, 0) \cup (0, 1]$.
• 1 mark — Case 1: $0 < \cos x \leq 1 \Rightarrow \sec x \geq 1$, with explicit inequality flip when taking reciprocals.
• 1 mark — Case 2: $-1 \leq \cos x < 0 \Rightarrow \sec x \leq -1$ (sign of reciprocal preserved; magnitude $\geq 1$).
• 1 mark — combines: range of $\sec x = (-\infty, -1] \cup [1, \infty)$ with endpoints achieved.
Part (c) — 2 marks
• 1 mark — correctly scales and shifts: $3 \sec x - 1$ has range $(-\infty, -4] \cup [2, \infty)$.
• 1 mark — identifies Trap 2 (sec's range excludes $(-1, 1)$); the claim is incorrect because the forbidden interval $(-1, 1)$ scales/shifts with the transformation but never disappears.
Your response:
Stuck on (b)? Sketching the graph of $\sec x$ helps visualise both cases.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Domain of $\tan 2x$ (3 marks)
Sample response. (a) $\tan 2x$ undefined when $2x = \frac{\pi}{2} + n\pi$, i.e. $x = \frac{\pi}{4} + \frac{n\pi}{2}$, $n \in \mathbb{Z}$. Domain: all real $x$ except $\mathbf{x = \frac{\pi}{4} + \frac{n\pi}{2}}$, $n \in \mathbb{Z}$.
(b) On $[0, \pi]$, choose $n = 0$: $x = \frac{\pi}{4}$. Choose $n = 1$: $x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$. So $x = \mathbf{\frac{\pi}{4}, \frac{3\pi}{4}}$.
Marking notes. 1 mark for the general formula (must include $n \in \mathbb{Z}$ — Trap 3). 1 mark for solving $2x = \frac{\pi}{2} + n\pi$ correctly. 1 mark for the two specific values on $[0, \pi]$. Common error: students give only $x = \frac{\pi}{4}$ (forget to iterate $n$) — costs 1 mark.
1.2 — Range of $4 - 2\cos x$ (3 marks)
Sample response. $-1 \leq \cos x \leq 1$. Multiply by $-2$ (reverses inequalities): $2 \geq -2 \cos x \geq -2$, i.e. $-2 \leq -2 \cos x \leq 2$. Add 4: $2 \leq 4 - 2 \cos x \leq 6$. Range = $\mathbf{[2, 6]}$.
Marking notes. 1 mark for starting from $\cos x \in [-1, 1]$. 1 mark for correctly handling the inequality flip on multiplying by $-2$. 1 mark for the final interval and explicit endpoints. Common error: ignoring the sign flip and writing $[2, 6]$ but via $-2 \leq 2 \cos x$ (wrong route) — mark only on final answer if working is otherwise sound.
1.3 — Domain and range of $\frac{1}{1 + \cos x}$ (4 marks)
Sample response. Domain: $1 + \cos x \neq 0 \Rightarrow \cos x \neq -1 \Rightarrow x \neq \pi + 2n\pi$, $n \in \mathbb{Z}$. Domain: all real $x$ except $x = \pi + 2n\pi$.
Range: $0 < 1 + \cos x \leq 2$ (since $-1 < \cos x \leq 1$ over the domain). Taking reciprocals of a positive inequality reverses order: $\frac{1}{1 + \cos x} \geq \frac{1}{2}$, with no upper bound (as $\cos x \to -1$, denominator $\to 0^+$, so the reciprocal $\to +\infty$). Range: $\left[\frac{1}{2}, \infty\right)$.
Marking notes. 1 mark for the domain restriction. 1 mark for the general-solution form. 1 mark for identifying $1 + \cos x \in (0, 2]$ over the domain. 1 mark for taking reciprocals correctly to obtain $\left[\frac{1}{2}, \infty\right)$. Common error: writing range $[\frac{1}{2}, \infty]$ (square bracket on $\infty$) costs 0.5.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Range of $\cos x$. Range is $[-1, 1]$. [1 mark] On the unit circle, $\cos x$ is the $x$-coordinate of the terminal point at angle $x$; every point on the unit circle has $x \in [-1, 1]$ (with both endpoints achieved at angles $0$ and $\pi$), so cosine attains every value in $[-1, 1]$ and no value outside. [1 mark]
Part (b) — Range of $\sec x$. Since $\sec x = \frac{1}{\cos x}$, we exclude $\cos x = 0$ (which would force division by zero), restricting attention to $\cos x \in [-1, 0) \cup (0, 1]$. [1 mark]
Case 1: $0 < \cos x \leq 1$. All quantities positive; taking reciprocals of a positive inequality reverses order: $\frac{1}{\cos x} \geq \frac{1}{1} = 1$. So $\sec x \geq 1$. [1 mark]
Case 2: $-1 \leq \cos x < 0$. All quantities negative; reciprocal preserves sign but flips magnitude order. Concretely: $|\cos x| \leq 1$ gives $|1/\cos x| \geq 1$, and the sign is negative, so $\sec x \leq -1$. [1 mark]
Combining both cases: range of $\sec x = (-\infty, -1] \cup [1, \infty)$, with endpoints $\pm 1$ achieved at $\cos x = \pm 1$ (i.e. $x = 0, \pi, 2\pi, \ldots$). [1 mark]
Part (c) — Range of $3 \sec x - 1$ and student claim. Apply the linear transformation to the range $(-\infty, -1] \cup [1, \infty)$: multiply by 3 to get $(-\infty, -3] \cup [3, \infty)$; subtract 1 to get $\mathbf{(-\infty, -4] \cup [2, \infty)}$. [1 mark]
The student claim is incorrect. It commits Trap 2 from the lesson: secant has a forbidden interval $(-1, 1)$ in its output, and any linear transformation $A \sec x + B$ preserves the forbidden interval (it just scales and shifts it). The presence of asymptotes makes the function unbounded in both directions, but it never fills in the gap — for $y = 3 \sec x - 1$ the gap is the open interval $(-4, 2)$, which no $x$ ever produces. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: (a) states range without justification; (b) gives the range but no case analysis; (c) wrong final interval (e.g. $[2, \infty)$ only). ≈ 3-4 marks.
Band 4: (a) unit-circle mentioned; (b) one case correctly argued; (c) correct interval but no Trap 2 link. ≈ 5-6 marks.
Band 5: Both cases derived in (b) but inequality-flip not made explicit; (c) correct and references Trap 2. ≈ 7 marks.
Band 6: Full rigour throughout: explicit unit-circle range in (a), explicit inequality flip in (b), and explicit gap-preservation argument in (c). 8/8.