Mathematics Advanced • Year 11 • Module 2 • Lesson 6
Reciprocal Trigonometric Functions
Apply reciprocal trig functions to real contexts: stellar parallax, surveying, mechanical advantage, simplifying compound expressions, and reading reciprocal graphs.
Problem 1 — Stellar parallax and secant (astronomical)
Astronomers measure the distance $d$ to a nearby star using the parallax angle $\theta$. For a star at distance $d$ parsecs, the geometric relationship reduces to $d = R \cdot \sec(90^\circ - p)$ for a baseline $R$ and parallax $p$. (See the lesson's "Why astronomers use parallax and secants" callout.) Take $R = 1$ astronomical unit and $p = \frac{\pi}{3}$ for this problem.
Set up: What are we solving for?
(i) Convert the expression $\sec(90^\circ - p)$ when $p = \frac{\pi}{3}$ into the reciprocal of a single basic trig function. 2 marks
(ii) Find the exact value of $\sec\left(\frac{\pi}{2} - \frac{\pi}{3}\right)$, then state $d$ in exact form. 2 marks
(iii) As the parallax $p$ approaches 0, what happens to $\sec\left(\frac{\pi}{2} - p\right)$, and what does that mean physically about the distance to a star with vanishingly small parallax? 2 marks
Stuck? Revisit lesson § Trap 3 — sec is undefined at odd multiples of $\frac{\pi}{2}$.Problem 2 — Inclined plane and cosecant (mechanical)
For a load on a frictionless inclined plane of angle $\theta$ to the horizontal, the mechanical advantage of using the slope (rather than lifting straight up) is $\text{MA} = \csc \theta$. A workshop ramp has $\theta = \frac{\pi}{6}$.
Set up: What are we solving for?
(i) Find the exact mechanical advantage at $\theta = \frac{\pi}{6}$. 2 marks
(ii) A second ramp uses $\theta = \frac{\pi}{4}$. Compute its MA, then state which ramp gives the greater MA and by what exact factor. 3 marks
(iii) Explain (referring to the range of $\csc$) why mechanical advantage on a frictionless ramp can never be less than 1. 2 marks
Problem 3 — Simplifying a compound expression (algebraic)
An engineering formula contains the expression $E(\theta) = \sin \theta \cdot \csc \theta + \cot \theta \cdot \tan \theta - \cos \theta \cdot \sec \theta$.
| Term | Rewrite as a fraction in sin / cos | Simplified value (where defined) |
|---|---|---|
| $\sin \theta \cdot \csc \theta$ | ||
| $\cot \theta \cdot \tan \theta$ | ||
| $\cos \theta \cdot \sec \theta$ |
Set up: What are we solving for?
(i) Fill the table above. 3 marks
(ii) Hence find $E(\theta)$ as a single number (where defined). 2 marks
(iii) State the values of $\theta$ in $[0, 2\pi]$ at which $E(\theta)$ is undefined, and identify which term forces each exclusion. 2 marks
Stuck? Revisit lesson § Worked Example — Simplifying with reciprocals.Problem 4 — Range detective (graphical)
The lesson states that the range of $\csc$ and $\sec$ is $(-\infty, -1] \cup [1, \infty)$, while the range of $\cot$ is all real numbers.
Set up: What are we solving for?
(i) For each value below, decide whether it can possibly equal $\sec \theta$ for some real $\theta$. Tick yes or no with one sentence of reasoning. 3 marks
A. $\sec \theta = \frac{1}{2}$ □ yes □ no Reason: ____________________
B. $\sec \theta = 2$ □ yes □ no Reason: ____________________
C. $\sec \theta = -\sqrt{2}$ □ yes □ no Reason: ____________________
(ii) A student writes "the range of $\cot \theta$ is $(-\infty, -1] \cup [1, \infty)$". Evaluate the claim in one sentence and give a counter-example. 2 marks
(iii) Find an angle $\theta$ in $[0, 2\pi]$ (exact form) such that $\csc \theta = \frac{2\sqrt{3}}{3}$. 2 marks
Problem 5 — Domain audit on a piecewise model
A signal-processing function is defined as $f(\theta) = \csc \theta + \sec \theta + \cot \theta$.
Set up: What are we solving for?
(i) List every value of $\theta$ in $[0, 2\pi]$ at which $f(\theta)$ is undefined, and state which term is responsible. 3 marks
$\theta = $ ________ — undefined because of ________
$\theta = $ ________ — undefined because of ________
$\theta = $ ________ — undefined because of ________
$\theta = $ ________ — undefined because of ________
(ii) Find the exact value of $f\left(\frac{\pi}{4}\right)$ and express it with a rational denominator. 2 marks
(iii) Write the general-form domain of $f$ (use $n \in \mathbb{Z}$) so the answer works for any real $\theta$, not just $[0, 2\pi]$. 2 marks
Stuck on (iii)? Combine the exclusions: $\theta \neq n\pi$ (from csc, cot) and $\theta \neq \frac{\pi}{2} + n\pi$ (from sec); merge to $\theta \neq \frac{n\pi}{2}$.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Stellar parallax
Set up. We're evaluating a secant of a complementary angle, then interpreting its asymptotic behaviour.
(i) $\sec\left(\frac{\pi}{2} - \frac{\pi}{3}\right) = \frac{1}{\cos(\pi/6)}$. (Equivalently, by the co-function identity, $\sec\left(\frac{\pi}{2} - \theta\right) = \csc \theta$, so this also equals $\csc \frac{\pi}{3}$.)
(ii) $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, so $\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. With $R = 1$ AU, $d = \mathbf{\frac{2\sqrt{3}}{3}}$ AU (about 1.155 AU).
(iii) As $p \to 0$, the argument $\frac{\pi}{2} - p \to \frac{\pi}{2}$, where $\cos$ vanishes and $\sec$ blows up to infinity. Physically: a star with no measurable parallax must be infinitely far away — consistent with Trap 3 (the asymptote at odd multiples of $\frac{\pi}{2}$).
Problem 2 — Inclined plane (csc)
Set up. We're computing cosecant at two reference angles, comparing them, and using the range of csc to argue a physical bound.
(i) $\csc \frac{\pi}{6} = \frac{1}{\sin \pi/6} = \frac{1}{1/2} = \mathbf{2}$.
(ii) $\csc \frac{\pi}{4} = \frac{1}{\sin \pi/4} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Since $2 > \sqrt{2}$, the first ramp ($\frac{\pi}{6}$) gives greater MA. Ratio: $\frac{2}{\sqrt{2}} = \mathbf{\sqrt{2}}$ — the first ramp is $\sqrt{2}$ times the second.
(iii) Range of $\csc$ is $(-\infty, -1] \cup [1, \infty)$, so for any acute $\theta$ (where csc is positive), $\csc \theta \geq 1$. Physically: the slope cannot give a smaller force than lifting straight up — the saving is bounded above 1 by the geometry.
Problem 3 — Compound simplification
Set up. We're collapsing each product to 1 (or relating to a reciprocal identity), summing/differencing, and listing all domain exclusions.
(i) $\sin \theta \cdot \csc \theta = \sin \theta \cdot \frac{1}{\sin \theta} = 1$. $\cot \theta \cdot \tan \theta = \frac{\cos \theta}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = 1$. $\cos \theta \cdot \sec \theta = \cos \theta \cdot \frac{1}{\cos \theta} = 1$.
(ii) $E(\theta) = 1 + 1 - 1 = \mathbf{1}$ where defined.
(iii) Undefined when $\sin \theta = 0$ (from $\csc, \cot$) OR $\cos \theta = 0$ (from $\sec, \tan$). On $[0, 2\pi]$: $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$. Causes: $\theta = 0, \pi, 2\pi$ break csc and cot; $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ break sec and tan.
Problem 4 — Range detective
Set up. We're applying the range constraint $|\sec|, |\csc| \geq 1$ and the unbounded range of cot.
(i) A: No — $|\sec \theta| \geq 1$ so $\frac{1}{2}$ is outside the range. B: Yes — e.g. $\sec \frac{\pi}{3} = 2$. C: Yes — e.g. $\sec \frac{3\pi}{4} = -\sqrt{2}$.
(ii) The claim is incorrect. $\cot \theta$ has range all real numbers (unlike csc and sec). Counter-example: $\cot \frac{\pi}{4} = 1$, but also $\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \approx 0.577$, which lies in $(-1, 1)$.
(iii) $\csc \theta = \frac{2\sqrt{3}}{3}$ means $\sin \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. So $\theta = \mathbf{\frac{\pi}{3}}$ or $\mathbf{\frac{2\pi}{3}}$.
Problem 5 — Domain audit
Set up. We're combining the exclusion sets of three reciprocal functions and evaluating $f$ at a clean reference angle.
(i) On $[0, 2\pi]$: $\theta = 0$ (csc & cot break), $\theta = \frac{\pi}{2}$ (sec breaks), $\theta = \pi$ (csc & cot break), $\theta = \frac{3\pi}{2}$ (sec breaks), $\theta = 2\pi$ (csc & cot break). Five exclusions if endpoints both listed.
(ii) $\csc \frac{\pi}{4} = \sqrt{2}$, $\sec \frac{\pi}{4} = \sqrt{2}$, $\cot \frac{\pi}{4} = 1$. Sum: $\sqrt{2} + \sqrt{2} + 1 = \mathbf{2\sqrt{2} + 1}$.
(iii) Domain: $\theta \in \mathbb{R}$ such that $\theta \neq \frac{n\pi}{2}$ for any $n \in \mathbb{Z}$. (This single condition merges $\theta \neq n\pi$ from csc/cot and $\theta \neq \frac{\pi}{2} + n\pi$ from sec.)