Mathematics Advanced • Year 11 • Module 2 • Lesson 8
Complementary Angle Relationships
Practise HSC-style writing on co-function identity proofs, complementary equation solving, and a structured derivation of the sin/cos identity from right-triangle geometry.
1. Short-answer questions
1.1 (a) Express $\sin 68^\circ$ as the cosine of a complementary angle. (b) Hence evaluate $\sin^2 68^\circ + \sin^2 22^\circ$ without a calculator. 3 marks Band 3-4
1.2 Solve $\tan 3\theta = \cot \theta$ for $0^\circ \leq \theta \leq 45^\circ$. Show all working. 3 marks Band 4
1.3 Prove that $\sec(90^\circ - \theta) = \csc \theta$ for all $\theta$ where both sides are defined. Explain why the qualifier "where both sides are defined" is essential, listing the values of $\theta$ in $[0^\circ, 360^\circ]$ where one or both sides fail. 4 marks Band 4-5
Stuck on 1.3? Both sides are defined whenever $\sin \theta \neq 0$.2. Extended response
2.1 A mathematics teacher wants a rigorous, multi-perspective derivation of the sine-cosine complementary identity for a high-achieving Year 11 cohort.
(a) Using a generic right-angled triangle with acute angles $\theta$ and $(90^\circ - \theta)$, prove from the SOH-CAH-TOA definitions that $\sin \theta = \cos(90^\circ - \theta)$ for every $\theta \in (0^\circ, 90^\circ)$. Identify which side swap drives the identity.
(b) Extend the result beyond the acute range using the unit-circle reflection across the line $y = x$, briefly explaining why $(\cos \theta, \sin \theta)$ reflected across $y = x$ becomes $(\sin \theta, \cos \theta) = (\cos(\frac{\pi}{2} - \theta), \sin(\frac{\pi}{2} - \theta))$.
(c) A student claims that, because $\sin \theta + \cos \theta = 1$ "should be the rule", $\sin 30^\circ + \cos 60^\circ = 1$. Evaluate the numerical claim, identify the structural misconception (referring to Trap 1 of the lesson), and write the correct co-function identity the student appears to have confused. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — clearly labels a right triangle with acute angles $\theta$ and $(90^\circ - \theta)$ and identifies opposite, adjacent, hypotenuse from $\theta$'s viewpoint.
• 1 mark — writes $\sin \theta = \frac{\text{opp}}{\text{hyp}}$ and observes that the side opposite $\theta$ is adjacent to $(90^\circ - \theta)$.
• 1 mark — concludes $\sin \theta = \frac{\text{adj}}{\text{hyp}}$ (from $(90^\circ - \theta)$'s viewpoint) $= \cos(90^\circ - \theta)$.
Part (b) — 3 marks
• 1 mark — states $(\cos \theta, \sin \theta)$ is the unit-circle point at angle $\theta$.
• 1 mark — correctly applies reflection across $y = x$ to swap coordinates, giving $(\sin \theta, \cos \theta)$.
• 1 mark — identifies this image as the point at angle $\frac{\pi}{2} - \theta$, giving $\cos(\frac{\pi}{2} - \theta) = \sin \theta$ for any real $\theta$ (extends beyond acute case).
Part (c) — 2 marks
• 1 mark — computes $\sin 30^\circ + \cos 60^\circ = \frac{1}{2} + \frac{1}{2} = 1$ (the numerical claim is "accidentally" correct here) but identifies that the rule $\sin + \cos = 1$ is not a general identity.
• 1 mark — explicitly names Trap 1 (mixing up "sine equals cosine of complement" with a false addition rule) and writes the correct identity: $\sin \theta = \cos(90^\circ - \theta)$ (or equivalently $\sin 30^\circ = \cos 60^\circ$).
Your response:
Stuck on (a)? Draw the triangle with the right angle bottom-left; place $\theta$ at the bottom-right vertex; label opp/adj/hyp from $\theta$'s viewpoint first.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Co-function + Identity 1 (3 marks)
Sample response. (a) $\sin 68^\circ = \cos(90^\circ - 68^\circ) = \mathbf{\cos 22^\circ}$.
(b) Using (a), $\sin^2 68^\circ = \cos^2 22^\circ$. So $\sin^2 68^\circ + \sin^2 22^\circ = \cos^2 22^\circ + \sin^2 22^\circ = \mathbf{1}$ (Pythagorean Identity 1).
Marking notes. 1 mark for (a). 1 mark for substituting the rewrite into (b). 1 mark for citing Identity 1 to reach the final answer.
1.2 — Solve $\tan 3\theta = \cot \theta$ (3 marks)
Sample response. Use $\cot \theta = \tan(90^\circ - \theta)$: $\tan 3\theta = \tan(90^\circ - \theta)$. Equating arguments (in $[0^\circ, 90^\circ]$ where tan is one-to-one within each branch): $3\theta = 90^\circ - \theta$, so $4\theta = 90^\circ$, giving $\theta = \mathbf{22.5^\circ}$. Within $[0^\circ, 45^\circ]$. ✓
Marking notes. 1 mark for converting $\cot \theta$ to $\tan(90^\circ - \theta)$. 1 mark for equating arguments and solving algebraically. 1 mark for the final answer and a check that it lies within the stated interval.
1.3 — Prove $\sec(90^\circ - \theta) = \csc \theta$ (4 marks)
Sample response. LHS = $\sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)} = \frac{1}{\sin \theta} = \csc \theta = $ RHS. ✓
Both sides defined iff $\sin \theta \neq 0$. On $[0^\circ, 360^\circ]$: $\sin \theta = 0$ at $\theta = 0^\circ, 180^\circ, 360^\circ$, so both sides fail at these three values. (Equivalently, $\cos(90^\circ - \theta) = 0$ when $90^\circ - \theta = \frac{\pi}{2} + n\pi$, i.e. $\theta = -n\pi$, giving the same set of exclusions.)
Marking notes. 1 mark for $\sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)}$. 1 mark for $\cos(90^\circ - \theta) = \sin \theta$. 1 mark for the conclusion $= \csc \theta$. 1 mark for the restriction list (must give the specific three values on $[0^\circ, 360^\circ]$, not just "$\sin \theta \neq 0$").
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Right-triangle derivation. Consider a right triangle ABC with the right angle at C. Let the angle at A be $\theta$; the angle at B is therefore $(90^\circ - \theta)$ (interior angles in a triangle sum to $180^\circ$ and one is $90^\circ$). Label the side opposite A as $a$ (this is also adjacent to B); opposite B as $b$ (also adjacent to A); hypotenuse $c$ (opposite C). [1 mark]
From $\theta$'s viewpoint: $\sin \theta = \frac{\text{opp to A}}{\text{hyp}} = \frac{a}{c}$. [1 mark]
From $(90^\circ - \theta)$'s viewpoint (angle B): $\cos(90^\circ - \theta) = \frac{\text{adj to B}}{\text{hyp}} = \frac{a}{c}$. (Side $a$ — opposite A — is adjacent to B.) Equating: $\sin \theta = \cos(90^\circ - \theta)$, for every acute $\theta$. [1 mark] The "side swap" driving the identity is: opposite to $\theta$ is the same physical side as adjacent to $(90^\circ - \theta)$.
Part (b) — Unit-circle extension. On the unit circle, the point at angle $\theta$ has coordinates $(\cos \theta, \sin \theta)$. [1 mark]
Reflection across the line $y = x$ swaps $x$- and $y$-coordinates, sending $(\cos \theta, \sin \theta) \mapsto (\sin \theta, \cos \theta)$. [1 mark]
Geometrically, this reflection is equivalent to mapping the angle $\theta$ to $\frac{\pi}{2} - \theta$ (the reflection of the angle ray across $y = x$). So the point at angle $\frac{\pi}{2} - \theta$ has coordinates $(\cos(\frac{\pi}{2} - \theta), \sin(\frac{\pi}{2} - \theta)) = (\sin \theta, \cos \theta)$. Matching components: $\cos(\frac{\pi}{2} - \theta) = \sin \theta$ and $\sin(\frac{\pi}{2} - \theta) = \cos \theta$, for every real $\theta$ — extending the result beyond the acute range. [1 mark]
Part (c) — Critique of the false rule. Numerical check: $\sin 30^\circ + \cos 60^\circ = \frac{1}{2} + \frac{1}{2} = 1$. The numerical claim is true here, but only because $\sin 30^\circ = \cos 60^\circ$ (complementary angles), so the sum $= 2 \sin 30^\circ = 1$. The claimed "rule" $\sin \theta + \cos \theta = 1$ fails for other values; e.g. $\sin 45^\circ + \cos 45^\circ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \neq 1$. [1 mark]
The student has committed Trap 1: confusing the co-function identity $\sin \theta = \cos(90^\circ - \theta)$ with a false addition rule. The correct identity is $\mathbf{\sin \theta = \cos(90^\circ - \theta)}$, equivalently $\sin 30^\circ = \cos 60^\circ$ (both equal $\frac{1}{2}$). The complementary structure is "swap function and shift to complement", not "add the two functions". [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: (a) writes the identity without derivation; (b) skipped or treated incorrectly; (c) numerical check only. ≈ 3-4 marks.
Band 4: (a) right-triangle reasoning present but side-labelling vague; (b) unit-circle mentioned without reflection argument; (c) names trap but no counter-example. ≈ 5-6 marks.
Band 5: (a) clear right-triangle proof; (b) reflection argument with sin/cos image identified; (c) counter-example given. ≈ 7 marks.
Band 6: Full rigour: labelled triangle with explicit side swap, $y=x$ reflection extended to general real $\theta$, and explicit articulation of Trap 1 with counter-example. 8/8.