Mathematics Advanced • Year 11 • Module 2 • Lesson 11
Graphs of Tangent and Cotangent
Practise HSC-style writing on transformed tangent/cotangent graphs, period derivations, and the conceptual link between the two functions.
1. Short-answer questions
1.1 Find the period and the equations of the vertical asymptotes (general form, $n \in \mathbb{Z}$) of $y = \tan\!\left(\dfrac{x}{3}\right)$. 3 marks Band 3
1.2 Consider $y = \cot(2x)$.
(a) State the period.
(b) Find the equations of the vertical asymptotes in $[0, \pi]$.
(c) Find the $x$-intercepts in $[0, \pi]$. 4 marks Band 3-4
1.3 Sketch $y = \tan(2x)$ for $0 \leq x \leq \pi$. On your sketch label every vertical asymptote (dashed) and every $x$-intercept (dot). State the period. 4 marks Band 4
Stuck on 1.3? Period of $\tan(2x)$ is $\pi/2$, so two full branch-pairs fit in $[0, \pi]$.2. Extended response
2.1 Consider the function $y = \tan x$ and its reciprocal-related partner $y = \cot x$.
(a) Derive algebraically, starting from $\tan x = \sin x / \cos x$, that $\tan(x + \pi) = \tan x$, and explain why this proves the period of tangent is $\pi$ rather than $2\pi$.
(b) Using the co-function identity $\cot x = \tan(\pi/2 - x)$, explain why the graph of $y = \cot x$ has the same shape as $y = \tan x$ but is reflected and shifted, then state explicitly the transformation.
(c) A student claims: "Tangent and cotangent are both undefined at $x = \pi/2$." Evaluate this claim. State for each function whether $\pi/2$ is in the domain, justify with the underlying ratio, and identify which of the lesson's three Traps the student has fallen into. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — uses $\sin(x+\pi) = -\sin x$ and $\cos(x+\pi) = -\cos x$ correctly.
• 1 mark — simplifies $(-\sin x)/(-\cos x) = \sin x / \cos x = \tan x$.
• 1 mark — explains that since $\pi$ is the smallest such shift (not $2\pi$), tangent's period is $\pi$.
Part (b) — 3 marks
• 1 mark — states $\cot x = \tan(\pi/2 - x)$.
• 1 mark — identifies $\pi/2 - x = -(x - \pi/2)$, splitting into a reflection in the $y$-axis and a horizontal shift of $\pi/2$.
• 1 mark — names the transformation explicitly: "reflect $y = \tan x$ in the $y$-axis, then shift right by $\pi/2$" (or equivalent).
Part (c) — 2 marks
• 1 mark — states correctly that $\tan(\pi/2)$ is undefined (cos zero) but $\cot(\pi/2)$ is defined and equals $0$ (sin = 1, cos = 0, ratio $0/1 = 0$).
• 1 mark — connects the student's error to Trap 03 (confusing tangent and cotangent asymptote locations).
Your response:
Stuck on (a)? Recall: $\sin(x+\pi) = -\sin x$ and $\cos(x+\pi) = -\cos x$ — both flips together preserve the ratio.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Period and asymptotes of $y = \tan(x/3)$ (3 marks)
Sample response. $b = 1/3$, so period $= \pi/(1/3) = $ $3\pi$ [1]. Asymptotes where $\cos(x/3) = 0 \Rightarrow x/3 = \pi/2 + n\pi \Rightarrow$ $x = 3\pi/2 + 3n\pi$ [2].
Marking notes. 1 mark for period (must show $\pi/b$, not $2\pi/b$). 2 marks for correct asymptote equation (1 for setting up $\cos(x/3) = 0$, 1 for solving). Common error: writing $x = \pi/2 + n\pi$ — failing to "un-do" the $\div 3$ inside the function.
1.2 — $y = \cot(2x)$ (4 marks)
Sample response.
(a) Period $= \pi/2$ [1].
(b) Asymptotes where $\sin(2x) = 0 \Rightarrow 2x = n\pi \Rightarrow x = n\pi/2$. In $[0, \pi]$: $x = 0,\ \pi/2,\ \pi$ [1].
(c) $x$-intercepts where $\cos(2x) = 0 \Rightarrow 2x = \pi/2 + n\pi \Rightarrow x = \pi/4 + n\pi/2$. In $[0, \pi]$: $x = \pi/4$ and $x = 3\pi/4$ [2].
Marking notes. (a) 1 mark. (b) 1 mark — all three asymptotes in the closed interval. (c) 2 marks — 1 for setting up correctly, 1 for both values. Common error in (c): students give $x = \pi/2$ (which is actually an asymptote, not an intercept).
1.3 — Sketch $y = \tan(2x)$ on $[0, \pi]$ (4 marks)
Sample response. Period $= \pi/2$ [1]. Asymptotes at $x = \pi/4,\ 3\pi/4$ [1]. $x$-intercepts at $x = 0,\ \pi/2,\ \pi$ [1]. Smooth increasing branches between consecutive asymptotes, passing through each intercept; curves never touch the dashed asymptote lines [1].
Marking notes. Award 1 mark per element: period stated, asymptotes drawn (dashed), intercepts marked, branches drawn correctly (smooth, increasing, approaching but not touching asymptotes — Trap 02 in the lesson). Loss of 1 mark for any branch drawn touching an asymptote.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Period of tangent. By definition, $\tan x = \sin x / \cos x$ wherever $\cos x \neq 0$. Then
$\tan(x + \pi) = \dfrac{\sin(x + \pi)}{\cos(x + \pi)} = \dfrac{-\sin x}{-\cos x} = \dfrac{\sin x}{\cos x} = \tan x$ [1 mark — using sin/cos shift identities; 1 mark — simplifying]
This shows $\tan$ is periodic with period dividing $\pi$. To rule out a smaller period, note that on $[0, \pi/2)$ the tangent function takes every real value exactly once (it's strictly increasing from $0$ to $+\infty$), so no value of $\tan$ on $(0, \pi/2)$ is repeated for any smaller shift. Hence the period is exactly $\pi$ — half the period of sine and cosine. [1 mark — period argument]
Part (b) — Cotangent as transformed tangent. The co-function identity gives $\cot x = \tan(\pi/2 - x)$ [1 mark]. Writing this as $\tan(-(x - \pi/2))$ separates two transformations: [1 mark — algebraic split]
1. Reflection in the $y$-axis (the inner minus sign), and 2. Horizontal translation $\pi/2$ to the right (the $-\pi/2$ inside). So the graph of $y = \cot x$ is the graph of $y = \tan x$ reflected in the $y$-axis and then shifted $\pi/2$ to the right. (Equivalently: reflected and shifted; alternative valid descriptions accepted.) [1 mark — explicit transformation]
Part (c) — Evaluating the claim. The claim is false. At $x = \pi/2$:
$\tan(\pi/2) = \sin(\pi/2)/\cos(\pi/2) = 1/0$ — undefined (vertical asymptote in $\tan$). However, $\cot(\pi/2) = \cos(\pi/2)/\sin(\pi/2) = 0/1 = 0$ — perfectly defined, in fact this is an $x$-intercept of $\cot$. [1 mark — distinguishing the two cases with the underlying ratios]
The student has fallen into Trap 03: confusing the asymptote locations of tangent (at $\pi/2 + n\pi$, where $\cos$ vanishes) with those of cotangent (at $n\pi$, where $\sin$ vanishes). The two functions have asymptotes at different places — they are not symmetric about $x = \pi/2$ in the way the student is imagining. [1 mark — Trap 03 named]
Total: 8/8.
Band descriptors for marker.
Band 3: Recites that tan has period $\pi$ but doesn't derive it; states $\cot = \tan(\pi/2 - x)$ without unpacking the transformation; identifies the claim is wrong but only partially. ≈ 3-4 marks.
Band 4: Derives part (a) correctly using sin/cos shifts; gives a partial transformation in (b); correctly answers (c) but doesn't name Trap 03. ≈ 5-6 marks.
Band 5: Complete derivation in (a) with period justification; correct transformation in (b); fully correct (c) with explicit Trap 03 reference. ≈ 7 marks.
Band 6: All of the above, with the additional rigour of proving in (a) that $\pi$ is the smallest such period (i.e. ruling out smaller candidates via the monotonicity of $\tan$ on one branch). 8/8.