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.