| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Prove trigonometric identity |
| Difficulty | Moderate -0.8 This is a straightforward algebraic manipulation of standard trig identities (converting tan to sin/cos, factoring) followed by a simple deduction. Part (i) requires routine algebraic steps with no conceptual difficulty, and part (ii) is immediate once (i) is established. This is easier than average for A-level, being a standard textbook-style identity proof with minimal problem-solving demand. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{s^2 - s^2}{c^2} \to \frac{s^2 - s^2c^2}{c^2} = \frac{s^2(1-c^2)}{c^2} \to t^2s^2\) | M1, M1, A1 | Use of \(s + c = t\); Use of \(s^2 + c^2 = 1\); All ok |
| (ii) RHS \(> 0 \to \tan^2 \theta > \sin^2 \theta\) QED; \(\tan \theta > \sin \theta\) if \(\theta\) acute | B1 | Realises RHS \(> 0\) |
$\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$
**(i)** $\frac{s^2 - s^2}{c^2} \to \frac{s^2 - s^2c^2}{c^2} = \frac{s^2(1-c^2)}{c^2} \to t^2s^2$ | M1, M1, A1 | Use of $s + c = t$; Use of $s^2 + c^2 = 1$; All ok
**(ii)** RHS $> 0 \to \tan^2 \theta > \sin^2 \theta$ QED; $\tan \theta > \sin \theta$ if $\theta$ acute | B1 | Realises RHS $> 0$\begin{enumerate}[label=(\roman*)]
\item Prove the identity $\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$. [3]
\item Use this result to explain why $\tan \theta > \sin \theta$ for $0° < \theta < 90°$. [1]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q1 [4]}}