| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 2 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Small angle approximation |
| Type | Geometric proof using small angles |
| Difficulty | Challenging +1.2 This requires setting up a right triangle (OC = cos θ, so AC = 1 - cos θ), then applying the small angle approximation cos θ ≈ 1 - θ²/2. It's a straightforward application of standard small angle results with basic trigonometry, but requires recognizing the geometric setup and making the connection—slightly above average difficulty due to the proof element and geometric reasoning needed. |
| Spec | 1.05e Small angle approximations: sin x ~ x, cos x ~ 1-x^2/2, tan x ~ x1.05g Exact trigonometric values: for standard angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = [AO - OC] = 1 - \cos\theta\) or \(\cos\theta = 1 - AC\) | M1 (1.1a) | AG; Allow \(AC = AO - OC\) with \(OC = \cos\theta\) for M1 |
| \(\theta\) small so \(AC\ 1 - \left(1 - \dfrac{\theta^2}{2}\right) = \dfrac{\theta^2}{2}\) | E1 (2.1) | Convincing completion |
**Question 3:**
$AC = [AO - OC] = 1 - \cos\theta$ or $\cos\theta = 1 - AC$ | M1 (1.1a) | AG; Allow $AC = AO - OC$ with $OC = \cos\theta$ for M1
$\theta$ small so $AC\ 1 - \left(1 - \dfrac{\theta^2}{2}\right) = \dfrac{\theta^2}{2}$ | E1 (2.1) | Convincing completion
---3 Fig. 3 shows a circle with centre O and radius 1 unit. Points A and B lie on the circle with angle $\mathrm { AOB } = \theta$ radians. C lies on AO , and BC is perpendicular to AO .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31bc8bde-8d37-4e97-94e2-e3e73aab55e9-4_648_627_1507_717}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Show that, when $\theta$ is small, $\mathrm { AC } \approx \frac { 1 } { 2 } \theta ^ { 2 }$.
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q3 [2]}}