Mathematics Advanced • Year 11 • Module 2 • Lesson 7
Pythagorean Identities
Practise HSC-style writing on Pythagorean identity proofs, full trig-value recovery, and a structured derivation of Identity 3 from first principles.
1. Short-answer questions
1.1 Prove $1 + \tan^2 \theta = \sec^2 \theta$ starting from $\sin^2 \theta + \cos^2 \theta = 1$. State any restrictions on $\theta$. 3 marks Band 3-4
1.2 Given $\tan \theta = -\frac{4}{3}$ and $\frac{3\pi}{2} < \theta < 2\pi$, find the exact values of $\sin \theta$ and $\cos \theta$. Show every step of the working. 4 marks Band 4
1.3 Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$ to a single trigonometric expression, justifying each step. State the restriction necessary for your simplification. 3 marks Band 4-5
Stuck on 1.3? Factor the numerator as a difference of squares using Identity 1.2. Extended response
2.1 A physics student is studying the elegant link between the three Pythagorean identities and wants a rigorous account of why all three follow from one geometric fact.
(a) Starting from the unit-circle equation $x^2 + y^2 = 1$, derive the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ using the substitutions $x = \cos \theta$ and $y = \sin \theta$. Justify why this is valid for every real $\theta$.
(b) Divide the fundamental identity by $\sin^2 \theta$ to derive Identity 3 ($1 + \cot^2 \theta = \csc^2 \theta$). State and justify the restriction this division imposes on $\theta$.
(c) Given $\cos \theta = -\frac{2}{3}$ and $\theta$ in Quadrant III, use Identity 3 (or, equivalently, the chain $\sin \theta \to \csc \theta \to \cot \theta$) to find $\cot \theta$ exactly. Then explain in 1-2 sentences why omitting the restriction "$\cos \theta \neq 0$" when dividing by $\cos^2 \theta$ (Trap 3 of the lesson) would be flagged by an HSC marker even though it would not change this particular numerical answer. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies $(\cos \theta, \sin \theta)$ as a point on the unit circle and substitutes into $x^2 + y^2 = 1$.
• 1 mark — justifies validity for all real $\theta$ (the unit circle parametrisation covers every angle).
Part (b) — 3 marks
• 1 mark — divides each term of Identity 1 by $\sin^2 \theta$ correctly.
• 1 mark — simplifies $\frac{\sin^2 \theta}{\sin^2 \theta} = 1$, $\frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$, $\frac{1}{\sin^2 \theta} = \csc^2 \theta$.
• 1 mark — states restriction $\sin \theta \neq 0$ (i.e. $\theta \neq n\pi$) and justifies it (division by zero would invalidate the step).
Part (c) — 3 marks
• 1 mark — finds $\sin^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$ and assigns negative sign by QIII.
• 1 mark — computes $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-2/3}{-\sqrt{5}/3} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$ (or equivalent route via $\csc \theta$).
• 1 mark — addresses Trap 3: a proof that omits the "$\cos \theta \neq 0$" restriction is incomplete because the division step is illegal there, even if the substituted numerical case happens to satisfy the restriction.
Your response:
Stuck on (a)? Draw the unit circle and mark $(\cos \theta, \sin \theta)$ explicitly.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Prove Identity 2 (3 marks)
Sample response. Start with $\sin^2 \theta + \cos^2 \theta = 1$. Divide every term by $\cos^2 \theta$ (provided $\cos \theta \neq 0$): $\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$. Simplify: $\tan^2 \theta + 1 = \sec^2 \theta$, i.e. $1 + \tan^2 \theta = \sec^2 \theta$. QED. Restriction: $\cos \theta \neq 0$, i.e. $\theta \neq \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$.
Marking notes. 1 mark for the division step. 1 mark for the simplification using $\frac{\sin}{\cos} = \tan$ and $\frac{1}{\cos} = \sec$. 1 mark for the explicit restriction. Common omission: forgetting the restriction costs the final mark even if the algebra is perfect.
1.2 — Find sin and cos from tan (4 marks)
Sample response. Use Identity 2: $\sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{16}{9} = \frac{25}{9}$, so $|\sec \theta| = \frac{5}{3}$. In QIV, $\sec \theta > 0$ (because $\cos > 0$), so $\sec \theta = \frac{5}{3}$ and $\cos \theta = \mathbf{\frac{3}{5}}$.
Then $\sin \theta = \tan \theta \cdot \cos \theta = -\frac{4}{3} \cdot \frac{3}{5} = \mathbf{-\frac{4}{5}}$.
Check: $\sin^2 \theta + \cos^2 \theta = \frac{16}{25} + \frac{9}{25} = 1$. ✓
Marking notes. 1 mark for substituting into Identity 2. 1 mark for computing $\sec^2$ and taking the square root. 1 mark for assigning sign by QIV. 1 mark for $\sin \theta$ via $\sin = \tan \cdot \cos$ (or equivalent). Including a numerical check strengthens the response but isn't required for full marks.
1.3 — Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$ (3 marks)
Sample response. Rewrite numerator using Identity 1: $\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$ (difference of squares). So $\frac{\sin^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} = \mathbf{1 + \cos \theta}$. Restriction: $\cos \theta \neq 1$, i.e. $\theta \neq 2n\pi$.
Marking notes. 1 mark for the Identity-1 rearrangement and difference-of-squares factorisation. 1 mark for the cancellation. 1 mark for the explicit restriction (must mention $\cos \theta \neq 1$ specifically; "where defined" alone is insufficient).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Derivation of Identity 1. Every point on the unit circle satisfies $x^2 + y^2 = 1$. By definition of sine and cosine on the unit circle, for any real $\theta$ the terminal point of the angle is $(\cos \theta, \sin \theta)$, with $x = \cos \theta$ and $y = \sin \theta$. [1 mark]
Substituting: $(\cos \theta)^2 + (\sin \theta)^2 = 1$, i.e. $\sin^2 \theta + \cos^2 \theta = 1$. The parametrisation $(\cos \theta, \sin \theta)$ covers every point on the unit circle as $\theta$ ranges over $\mathbb{R}$, so the identity holds for all real $\theta$. [1 mark]
Part (b) — Derivation of Identity 3. Take Identity 1 and divide every term by $\sin^2 \theta$ (provided $\sin \theta \neq 0$, otherwise division by zero): $\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}$. [1 mark]
Simplify each term: $1 + \cot^2 \theta = \csc^2 \theta$. [1 mark]
Restriction: $\sin \theta \neq 0$, i.e. $\theta \neq n\pi$ for $n \in \mathbb{Z}$. This restriction is necessary because the division step would otherwise involve $\frac{1}{0}$, which is undefined; without it, the derivation is not mathematically valid even though the final formula has the same form. [1 mark]
Part (c) — Computing $\cot \theta$ and addressing Trap 3. From Identity 1: $\sin^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$. In QIII sine is negative, so $\sin \theta = -\frac{\sqrt{5}}{3}$. [1 mark]
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-2/3}{-\sqrt{5}/3} = \frac{2}{\sqrt{5}} = \mathbf{\frac{2\sqrt{5}}{5}}$ (rationalised). Both negatives cancel, consistent with $\cot > 0$ in QIII (ASTC: "T" for tan, hence cot, is positive). [1 mark]
Even though our specific $\cos \theta = -\frac{2}{3} \neq 0$ satisfies the restriction, an HSC proof of Identity 2 that omits "$\cos \theta \neq 0$" when dividing is structurally incomplete (Trap 3 in the lesson). The proof must be valid for all values it claims to cover; the restriction defines that scope. Markers deduct for missing restrictions because they reveal logical gaps in the argument, not just minor cosmetic oversights. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: Quotes Identity 1 in (a) without unit-circle derivation; derives Identity 3 in (b) without stating the restriction; computes $\cot \theta$ in (c) with sign error or no quadrant analysis. ≈ 3-4 marks.
Band 4: (a) mentions unit circle vaguely; (b) restriction stated but not justified; (c) correct numerical answer but no Trap 3 commentary. ≈ 5-6 marks.
Band 5: All three parts technically correct; (c) mentions Trap 3 but doesn't fully articulate the structural gap. ≈ 7 marks.
Band 6: Full rigour throughout: explicit parametrisation justification in (a), restriction + reason in (b), quadrant-based sign with rationalisation in (c), and the structural argument about restrictions in proofs. 8/8.