| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (i) requires expanding and applying the sum of cubes factorization (a³+b³=(a+b)(a²-ab+b²)), which is a standard algebraic identity. Part (ii) involves substituting the proven identity and solving a cubic equation that factors nicely into sin θ = cos θ, yielding straightforward angle solutions. This is a routine multi-step question testing standard identities and equation-solving techniques with no novel insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) \equiv \sin^3\theta + \cos^3\theta\) | Accept abbreviations s and c | |
| LHS \(= \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta\) | M1 | Expansion |
| \(= \sin\theta(1 - \cos^2\theta) + \cos\theta(1 - \sin^2\theta)\) or \((s + c - c(1-c^2) - s(1-s^2))\) | M1A1 | Uses identity twice. Everything correct. AG |
| Uses \(\sin^2\theta + \cos^2\theta = 1 \rightarrow \sin^3\theta + \cos^3\theta\) (RHS) | or from RHS: M1 for use of trig ID twice | |
| Or | ||
| LHS \(= (\sin\theta + \cos\theta)(\sin^2\theta + \cos^2\theta - \sin\theta\cos\theta)\) | M1 | M1 for factorisation |
| \(= \sin^3\theta + \sin\theta\cos^2\theta - \sin^2\theta\cos\theta + \cos\theta\sin^2\theta + \cos^3\theta - \sin\theta\cos^2\theta = \sin^3\theta + \cos^3\theta\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) = 3\cos^3\theta \rightarrow \sin^3\theta = 2\cos^3\theta\) | M1 | |
| \(\rightarrow \tan^3\theta = 2 \rightarrow \theta = 51.6°\) or \(231.6°\) (only) | A1A1FT | Uses \(\tan^3\theta = \sin^3\theta \div \cos^3\theta\). A1 CAO. A1FT, \(180 +\) their acute angle. \(\tan^3\theta = 0\) gets M0 |
## Question 4(i):
| $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) \equiv \sin^3\theta + \cos^3\theta$ | | Accept abbreviations s and c |
|---|---|---|
| LHS $= \sin\theta + \cos\theta - \sin^2\theta\cos\theta - \sin\theta\cos^2\theta$ | M1 | Expansion |
| $= \sin\theta(1 - \cos^2\theta) + \cos\theta(1 - \sin^2\theta)$ or $(s + c - c(1-c^2) - s(1-s^2))$ | M1A1 | Uses identity twice. Everything correct. AG |
| Uses $\sin^2\theta + \cos^2\theta = 1 \rightarrow \sin^3\theta + \cos^3\theta$ (RHS) | | or from RHS: M1 for use of trig ID twice |
| **Or** | | |
| LHS $= (\sin\theta + \cos\theta)(\sin^2\theta + \cos^2\theta - \sin\theta\cos\theta)$ | M1 | M1 for factorisation |
| $= \sin^3\theta + \sin\theta\cos^2\theta - \sin^2\theta\cos\theta + \cos\theta\sin^2\theta + \cos^3\theta - \sin\theta\cos^2\theta = \sin^3\theta + \cos^3\theta$ | M1A1 | |
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## Question 4(ii):
| $(\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta) = 3\cos^3\theta \rightarrow \sin^3\theta = 2\cos^3\theta$ | M1 | |
|---|---|---|
| $\rightarrow \tan^3\theta = 2 \rightarrow \theta = 51.6°$ or $231.6°$ (only) | A1A1FT | Uses $\tan^3\theta = \sin^3\theta \div \cos^3\theta$. A1 CAO. A1FT, $180 +$ their acute angle. $\tan^3\theta = 0$ gets M0 |
---4 (i) Prove the identity $( \sin \theta + \cos \theta ) ( 1 - \sin \theta \cos \theta ) \equiv \sin ^ { 3 } \theta + \cos ^ { 3 } \theta$.\\
(ii) Hence solve the equation $( \sin \theta + \cos \theta ) ( 1 - \sin \theta \cos \theta ) = 3 \cos ^ { 3 } \theta$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q4 [6]}}