Mathematics Advanced • Year 11 • Module 2 • Lesson 11
Graphs of Tangent and Cotangent
Apply tangent and cotangent graphs to scenarios where asymptotes carry physical meaning — lighthouse beams, surveying angles, sundial shadows, electronic oscillators, and pendulum tilt.
Problem 1 — Lighthouse beam striking a coast (geometric)
A lighthouse stands $200\,\text{m}$ inland from a long straight coastline. A rotating beam sweeps from the lighthouse with angle $\theta$ measured from the perpendicular to the coast (so $\theta = 0$ points straight at the nearest point on the coast). The distance $d$ (in metres) from this nearest point to the spot where the beam hits the coast satisfies $d = 200 \tan \theta$.
Set up: What are we solving for?
(i) Find $d$ when $\theta = \pi/4$ (i.e. the beam is at $45^\circ$). 2 marks
(ii) As $\theta \to \pi/2^-$, what happens to $d$? Explain how this corresponds to a feature of the graph of $y = \tan x$. 2 marks
(iii) A range cut-off forces $|d| \leq 1000\,\text{m}$. Find the maximum value of $\theta$ (in exact radians or to 3 dp) at which the beam still lands on the cut-off section of coast. 2 marks
Stuck? The asymptote at $\theta = \pi/2$ corresponds to the beam pointing parallel to the coast, where it never lands.Problem 2 — Sundial shadow length (cotangent model)
A vertical pole of height $1\,\text{m}$ casts a shadow of length $L$ (in metres) when the sun is at altitude $\alpha$ above the horizon. Then $L = \cot \alpha$.
Set up: What are we solving for?
(i) Find $L$ exactly when $\alpha = \pi/3$ (sun is $60^\circ$ up) and when $\alpha = \pi/4$ (sun at $45^\circ$). 2 marks
(ii) As $\alpha \to 0^+$ (sun just above horizon at sunset), what happens to $L$? Which feature of the graph of $y = \cot x$ is this? 2 marks
(iii) The sundial face only fits shadows up to $L = 3\,\text{m}$. What is the smallest altitude $\alpha$ (in exact form, then to 3 dp) for which the shadow still fits? 3 marks
Problem 3 — Electronic oscillator output (compressed period)
An electronic test generator produces a waveform $V = 5 \tan(100\pi t)$ volts, where $t$ is in seconds. (This is unusual — most circuits use sine — but tangent is occasionally used in timing analysis.)
Set up: What are we solving for?
(i) Find the period of the waveform in seconds. 2 marks
(ii) Find the times $t$ (in seconds, in the first period $t \in [0, T)$) where the voltage becomes undefined (i.e. asymptote locations). 3 marks
(iii) In an ideal asymptote-free model, what value would the circuit's voltage approach as $t \to$ (your answer in part ii)$^-$? Real circuits clip at $\pm 12\,\text{V}$ — write a one-sentence explanation of why this is necessary. 2 marks
Stuck? Period of $\tan(bx)$ is $\pi/b$, and asymptotes occur where $\cos(bx) = 0$.Problem 4 — Surveying with elevation angles
A surveyor stands a horizontal distance $50\,\text{m}$ from the base of a cliff. She measures the elevation angle $\theta$ to the top of the cliff. The cliff height is $h = 50 \tan \theta$.
Set up: What are we solving for?
(i) Complete the table of heights (use exact form where possible). 3 marks (1 each)
| $\theta$ (radians) | $\tan \theta$ (exact) | $h = 50\tan\theta$ (metres) |
|---|---|---|
| $\pi/6$ | ||
| $\pi/4$ | ||
| $\pi/3$ |
(ii) The surveyor's clinometer cannot read angles within $0.1\,\text{rad}$ of $\pi/2$. What is the largest height the instrument can reliably measure (give to nearest metre)? 2 marks
(iii) Suppose the cliff is $200\,\text{m}$ tall. Find the exact elevation angle $\theta$ she records, then express this angle in degrees to 1 dp. 2 marks
Problem 5 — Pendulum tilt and cotangent model
A pendulum of length $1\,\text{m}$ hangs from a fixed pivot. When tilted by angle $\theta$ from vertical (with $0 < \theta < \pi$), the horizontal displacement of the bob is $x = \sin \theta$ metres and its height below the pivot is $y = \cos \theta$ metres. The ratio $y/x = \cot \theta$ describes "how far below the pivot per unit horizontal swing".
Set up: What are we solving for?
(i) Find $\cot \theta$ when $\theta = \pi/6$ and when $\theta = \pi/3$ (exact form). 2 marks
(ii) What is $\cot(\pi/2)$? Interpret geometrically. 2 marks
(iii) A student claims $\cot \theta > 0$ for all angles $\theta \in (0, \pi)$. Evaluate this claim. State the interval(s) on $(0, \pi)$ where $\cot \theta$ is positive and where it is negative, and the single $\theta$ where it equals zero. 2 marks
Stuck on (iii)? $\cot \theta = \cos \theta / \sin \theta$. On $(0, \pi)$, $\sin \theta > 0$ throughout, so the sign of $\cot \theta$ follows $\cos \theta$.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Lighthouse
Set up. Evaluating $d = 200 \tan\theta$ at specific angles, then interpreting the asymptote and inverting for $\theta$.
(i) $d = 200 \tan(\pi/4) = 200 \times 1 = $ $200\,\text{m}$.
(ii) As $\theta \to \pi/2^-$, $\tan\theta \to +\infty$, so $d \to +\infty$. This is the vertical asymptote of $y = \tan x$ at $x = \pi/2$. Physically, the beam runs parallel to the coast and never strikes it.
(iii) Solve $200 \tan\theta = 1000 \Rightarrow \tan\theta = 5 \Rightarrow \theta = \arctan 5 \approx$ $1.373\,\text{rad}$ (about $78.69^\circ$).
Problem 2 — Sundial
Set up. Computing $L = \cot \alpha$ at exact angles, then inverting.
(i) $\cot(\pi/3) = \cos(\pi/3)/\sin(\pi/3) = (1/2)/(\sqrt{3}/2) = $ $1/\sqrt{3} = \sqrt{3}/3 \approx 0.577\,\text{m}$. $\cot(\pi/4) = 1/1 = $ $1\,\text{m}$.
(ii) As $\alpha \to 0^+$, $\sin\alpha \to 0^+$ while $\cos\alpha \to 1$, so $\cot\alpha \to +\infty$ and $L \to +\infty$. This is the vertical asymptote of $\cot x$ at $x = 0$ — physically, the shadow stretches without bound as the sun sets.
(iii) Solve $\cot\alpha = 3 \Rightarrow \tan\alpha = 1/3 \Rightarrow \alpha = \arctan(1/3)$. Exact form: $\alpha = \arctan(1/3)$; decimal $\approx$ $0.322\,\text{rad}$ ($\approx 18.43^\circ$).
Problem 3 — Electronic oscillator
Set up. Period and asymptotes of $V = 5\tan(100\pi t)$.
(i) Period $= \pi/(100\pi) = $ $1/100 = 0.01\,\text{s}$ (i.e. 10 ms).
(ii) Asymptotes where $\cos(100\pi t) = 0 \Rightarrow 100\pi t = \pi/2 + n\pi \Rightarrow t = 1/200 + n/100$. In first period $[0, 0.01)$: $t = 1/200\,\text{s} = 0.005\,\text{s}$ (5 ms).
(iii) As $t \to (0.005)^-$, ideal $V \to +\infty$. Clipping at $\pm 12\,\text{V}$ is necessary because real components (op-amps, supply rails) can only output voltages within their supply range — without clipping the circuit would either saturate destructively or draw unbounded current.
Problem 4 — Surveying with elevation angles
Set up. Table evaluation, then inverting for $\theta$ near the asymptote.
(i) Table: $\pi/6$: $\tan = 1/\sqrt{3}$, $h = 50/\sqrt{3} = $ $50\sqrt{3}/3 \approx 28.87\,\text{m}$. $\pi/4$: $\tan = 1$, $h = $ $50\,\text{m}$. $\pi/3$: $\tan = \sqrt{3}$, $h = $ $50\sqrt{3} \approx 86.60\,\text{m}$.
(ii) Largest reliable $\theta = \pi/2 - 0.1 \approx 1.4708$ rad. $h_{\max} = 50 \tan(1.4708) \approx 50 \times 9.967 \approx$ $498\,\text{m}$.
(iii) $200 = 50\tan\theta \Rightarrow \tan\theta = 4 \Rightarrow \theta = \arctan 4$ ≈ $1.3258$ rad ≈ $75.96^\circ$. (Note: well within the instrument's reliable range.)
Problem 5 — Pendulum tilt
Set up. Evaluating $\cot \theta$ at specific exact angles and analysing its sign across the domain $(0, \pi)$.
(i) $\cot(\pi/6) = \cos(\pi/6)/\sin(\pi/6) = (\sqrt{3}/2)/(1/2) = $ $\sqrt{3}$. $\cot(\pi/3) = (1/2)/(\sqrt{3}/2) = $ $1/\sqrt{3} = \sqrt{3}/3$.
(ii) $\cot(\pi/2) = \cos(\pi/2)/\sin(\pi/2) = 0/1 = $ $0$. Geometrically the bob is horizontal from the pivot at $\theta = \pi/2$, so $y = 0$ (no height below pivot) and the ratio $y/x = 0$.
(iii) The claim is incorrect. On $(0, \pi)$, $\sin\theta > 0$, but $\cos\theta > 0$ on $(0, \pi/2)$ and $\cos\theta < 0$ on $(\pi/2, \pi)$. So $\cot\theta > 0$ on $(0, \pi/2)$, $\cot\theta < 0$ on $(\pi/2, \pi)$, and $\cot(\pi/2) = 0$ exactly.