| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Standard +0.8 This question requires proving a triple angle identity using compound angle formulas, algebraic manipulation to deduce a second identity, then solving a non-standard trigonometric equation by substitution. While the proof is methodical, the final equation requires recognizing that cot²θ = cos²θ/sin²θ and solving 4sin³θ = cos²θ, which involves substitution and solving a cubic. This goes beyond routine A-level exercises but remains accessible with solid technique. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sin 3\theta = \sin(2\theta + \theta)\) | M1 | Attempt \(\sin(2\theta + \theta)\). Needs two terms but may be incorrect |
| \(= \sin 2\theta \cos \theta + \cos 2\theta \sin \theta\) | ||
| \(= 2\sin \theta \cos^2 \theta + (1 - 2\sin^2 \theta) \sin \theta\) | M1 | Use any \(\cos 2\theta\) or \(\sin 2\theta\) identity |
| \(= 2\sin \theta (1 - \sin^2 \theta) + (1 - 2\sin^2 \theta) \sin \theta\) | M1 | Use \(\cos^2 \theta = 1 - \sin^2 \theta\) anywhere |
| \(= 3\sin \theta - 4\sin^3 \theta\) | A1 | Obtain the AG NIS |
| \(\sin \theta + \sin 3\theta\) | B1 | Factorise 4sin \(\theta\) or sin \(\theta\) and correctly obtain the AG |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(\frac{\cos^2 \theta}{\sin^2 \theta} = 4\sin\theta \cos^2 \theta\) | M1 | Identify \(\frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta\) |
| \(\frac{1}{4} = \sin^2 \theta\) | M1 | Attempt to solve for \(\theta\) |
| \(\theta = 39.0\) or \(141.0\) | A1 | Obtain 39.0 and 141.0 |
| B1 | [9] | Recognise \(\cos^2 \theta = 0\) and state \(90°\) |
(i) $\sin 3\theta = \sin(2\theta + \theta)$ | M1 | Attempt $\sin(2\theta + \theta)$. Needs two terms but may be incorrect
$= \sin 2\theta \cos \theta + \cos 2\theta \sin \theta$ |
$= 2\sin \theta \cos^2 \theta + (1 - 2\sin^2 \theta) \sin \theta$ | M1 | Use any $\cos 2\theta$ or $\sin 2\theta$ identity
$= 2\sin \theta (1 - \sin^2 \theta) + (1 - 2\sin^2 \theta) \sin \theta$ | M1 | Use $\cos^2 \theta = 1 - \sin^2 \theta$ anywhere
$= 3\sin \theta - 4\sin^3 \theta$ | A1 | Obtain the AG NIS
$\sin \theta + \sin 3\theta$ | B1 | Factorise 4sin $\theta$ or sin $\theta$ and correctly obtain the AG
$= 4\sin \theta - 4\sin^3 \theta$
$= 4\sin \theta (1 - \sin^2 \theta)$
$= 4\sin \theta \cos^2 \theta$
(ii) $\frac{\cos^2 \theta}{\sin^2 \theta} = 4\sin\theta \cos^2 \theta$ | M1 | Identify $\frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$
$\frac{1}{4} = \sin^2 \theta$ | M1 | Attempt to solve for $\theta$
$\theta = 39.0$ or $141.0$ | A1 | Obtain 39.0 and 141.0
| B1 | [9] | Recognise $\cos^2 \theta = 0$ and state $90°$\begin{enumerate}[label=(\roman*)]
\item Prove that $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$ and deduce that
$$\sin \theta + \sin 3\theta = 4 \sin \theta \cos^2 \theta.$$ [5]
\item Hence find the values of $\theta$ such that $0° < \theta < 180°$ that satisfy the equation
$$\cot^2 \theta = \sin \theta + \sin 3\theta.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q9 [9]}}