Mathematics Advanced • Year 11 • Module 2 • Lesson 6
Reciprocal Trigonometric Functions
Practise HSC-style writing on reciprocal trig evaluation, identity-driven simplification, and a structured derivation of the range of cosecant.
1. Short-answer questions
1.1 (a) Find the exact value of $\sec \frac{7\pi}{6}$. (b) Find the exact value of $\cot \frac{5\pi}{4}$. Rationalise where appropriate. 3 marks Band 3
1.2 Simplify $\frac{\sec \theta}{\tan \theta}$ to a single trigonometric function, showing every algebraic step. State all restrictions on $\theta$. 3 marks Band 3-4
1.3 Given $\cos \theta = -\frac{3}{5}$ and $\theta$ in Quadrant II, find the exact values of $\sec \theta$, $\csc \theta$ and $\cot \theta$. Show working for the latter two. 4 marks Band 4
Stuck on 1.3? Use $\sin^2 \theta + \cos^2 \theta = 1$ to find sin, then assign sign by quadrant.2. Extended response
2.1 A control engineer uses the function $g(\theta) = \csc \theta$ to model a mechanical-advantage curve for a tilting platform, with $\theta$ in radians representing the platform angle.
(a) Starting from the reciprocal definition and the known range of $\sin \theta$, derive the range of $\csc \theta$ over $\theta \in \mathbb{R}$ where defined. Justify each step of the inequality manipulation.
(b) State the smallest positive value of $\theta$ at which $g(\theta) = -2$, giving your answer exactly. Show working using quadrant analysis (ASTC).
(c) Explain in 1-2 sentences why $g(\theta) = \frac{1}{2}$ has no solution, referring directly to your range derivation in (a). Then explain why the engineer's mechanical model would treat $\theta = n\pi$ as a "forbidden configuration" of the platform, citing the lesson's Trap 3 about domain restrictions. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — states $\sin \theta \in [-1, 1]$ excluding 0, with reciprocal definition $\csc \theta = \frac{1}{\sin \theta}$.
• 1 mark — correctly inverts the inequalities in both positive and negative branches (handles the sign flip).
• 1 mark — concludes range is $(-\infty, -1] \cup [1, \infty)$ with both endpoints included.
Part (b) — 3 marks
• 1 mark — recognises $\csc \theta = -2 \Rightarrow \sin \theta = -\frac{1}{2}$.
• 1 mark — locates negative sin in QIII or QIV with reference angle $\frac{\pi}{6}$.
• 1 mark — identifies the smallest positive solution as $\frac{7\pi}{6}$ (QIII first because $\frac{7\pi}{6} < \frac{11\pi}{6}$).
Part (c) — 2 marks
• 1 mark — ties the impossibility of $\csc \theta = \frac{1}{2}$ directly to the range derivation (no value of csc in $(-1, 1)$).
• 1 mark — states that $\theta = n\pi$ makes $\sin \theta = 0$, so $\csc \theta$ is undefined — the platform model has no defined output (Trap 3: domain must exclude every odd multiple of $\frac{\pi}{2}$ for sec; analogously every integer multiple of $\pi$ for csc).
Your response:
Stuck on (a)? Split into two cases: $0 < \sin \theta \leq 1$ and $-1 \leq \sin \theta < 0$, taking reciprocals (which flip inequalities when dividing through positive vs negative).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Reciprocal evaluations (3 marks)
Sample response. (a) $\frac{7\pi}{6}$ is in QIII, reference $\frac{\pi}{6}$. $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$; cos negative in QIII, so $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$. $\sec \frac{7\pi}{6} = \frac{1}{-\sqrt{3}/2} = -\frac{2}{\sqrt{3}} = \mathbf{-\frac{2\sqrt{3}}{3}}$. (b) $\frac{5\pi}{4}$ in QIII, reference $\frac{\pi}{4}$. $\tan \frac{\pi}{4} = 1$; tan positive in QIII, so $\tan \frac{5\pi}{4} = 1$. $\cot \frac{5\pi}{4} = \mathbf{1}$.
Marking notes. 1.5 each. Award 0.5 for quadrant/reference, 0.5 for sign, 0.5 for rationalised exact form. Common error: dropping rationalisation on (a) (writing $-\frac{2}{\sqrt{3}}$ as final) costs 0.5.
1.2 — $\frac{\sec \theta}{\tan \theta}$ (3 marks)
Sample response. $\frac{\sec \theta}{\tan \theta} = \frac{1/\cos \theta}{\sin \theta / \cos \theta} = \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} = \mathbf{\csc \theta}$. Restrictions: $\cos \theta \neq 0$ and $\sin \theta \neq 0$, so $\theta \neq \frac{n\pi}{2}$ for any $n \in \mathbb{Z}$.
Marking notes. 1 mark for converting both reciprocals to sin/cos form. 1 mark for the cancellation and identification as $\csc \theta$. 1 mark for the complete restriction set (must mention both $\sin \theta \neq 0$ and $\cos \theta \neq 0$ — one alone forfeits this mark).
1.3 — Reciprocals from $\cos \theta = -\frac{3}{5}$, QII (4 marks)
Sample response. $\sec \theta = \frac{1}{\cos \theta} = \mathbf{-\frac{5}{3}}$.
Find $\sin \theta$ from Pythagoras: $\sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$, so $|\sin \theta| = \frac{4}{5}$. In QII, sin positive, so $\sin \theta = \frac{4}{5}$.
$\csc \theta = \frac{1}{\sin \theta} = \mathbf{\frac{5}{4}}$.
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-3/5}{4/5} = \mathbf{-\frac{3}{4}}$.
Marking notes. 1 mark for $\sec \theta$. 1 mark for using Pythagoras to find $|\sin \theta|$. 1 mark for assigning the correct sign by quadrant. 1 mark for the final $\csc$ and $\cot$ values (both required for the mark).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Derivation of the range of $\csc \theta$. The function $\sin \theta$ has range $[-1, 1]$, but at $\sin \theta = 0$ the reciprocal $\csc \theta = \frac{1}{\sin \theta}$ is undefined, so we work with $\sin \theta \in [-1, 0) \cup (0, 1]$. [1 mark]
Case 1: $0 < \sin \theta \leq 1$. Taking reciprocals of a positive inequality reverses the order, giving $\frac{1}{\sin \theta} \geq \frac{1}{1} = 1$, so $\csc \theta \geq 1$.
Case 2: $-1 \leq \sin \theta < 0$. The reciprocal of a negative quantity is negative; here $|\sin \theta| \leq 1$ gives $|1/\sin \theta| \geq 1$, so $\csc \theta \leq -1$. [1 mark]
Combining both cases: range of $\csc \theta$ is $(-\infty, -1] \cup [1, \infty)$, with endpoints achieved at $\sin \theta = \pm 1$ (i.e. $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$, $\ldots$). [1 mark]
Part (b) — Smallest positive $\theta$ with $g(\theta) = -2$. $\csc \theta = -2 \Leftrightarrow \sin \theta = -\frac{1}{2}$. [1 mark]
Sine is negative in QIII and QIV. Reference angle: $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$. [1 mark] Candidate solutions in $[0, 2\pi]$: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ (QIII), and $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ (QIV). The smallest positive is $\mathbf{\frac{7\pi}{6}}$. [1 mark]
Part (c) — No solution to $g(\theta) = \frac{1}{2}$, and forbidden configurations. The value $\frac{1}{2}$ lies in $(-1, 1)$, which is excluded from the range derived in (a); no $\theta$ produces $\csc \theta = \frac{1}{2}$. [1 mark]
$\theta = n\pi$ makes $\sin \theta = 0$, so $\csc \theta = \frac{1}{0}$ is undefined — the model has no output, mirroring the lesson's Trap 3 about excluding zeros of the denominator function. The engineer should treat platform angles at integer multiples of $\pi$ as forbidden because the reciprocal model produces a singularity. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: States the range without derivation in (a); solves $\sin \theta = -\frac{1}{2}$ but gives only one solution or wrong quadrant; vague handling of (c). ≈ 3-4 marks.
Band 4: Derivation in (a) covers one case only (positive sin); (b) lists both candidate solutions but doesn't identify "smallest positive"; (c) mentions undefined points without referencing the range. ≈ 5-6 marks.
Band 5: Both cases derived in (a); (b) clean; (c) connects to derivation and identifies $\theta = n\pi$ correctly but doesn't explicitly reference Trap 3. ≈ 7 marks.
Band 6: Full rigour, explicit reciprocal inequality manipulation in both signs, clean QIII identification, and explicit Trap 3 link. 8/8.