| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Small angle approximation |
| Type | Derive related approximation formula |
| Difficulty | Standard +0.8 Part (a) is trivial recall of a standard trig identity. Part (b) requires algebraic manipulation to transform a given approximation formula by substituting x+π/2 for x, then simplifying the resulting rational expression—this involves careful algebraic work but follows a clear path once the substitution is identified. The question is moderately challenging due to the algebraic complexity and the need to work with an unfamiliar approximation formula, but it's essentially guided manipulation rather than requiring deep insight or novel problem-solving. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \sin\!\left(x + \dfrac{\pi}{2}\right)\) is a translation of \(y = \sin x\) | M1 | 3.1a |
| \(\dfrac{\pi}{2}\) to left and so is the same as \(y = \cos x\) | E1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \([\cos x \approx]\dfrac{16\!\left(x+\dfrac{\pi}{2}\right)\!\left(\dfrac{\pi}{2}-x\right)}{5\pi^2 - 4\!\left(x+\dfrac{\pi}{2}\right)\!\left(\dfrac{\pi}{2}-x\right)}\) | M1* | 2.1 |
| \([\cos x \approx]\dfrac{16\!\left(\dfrac{\pi^2}{4} - x^2\right)}{5\pi^2 - 4\!\left(\dfrac{\pi^2}{4} - x^2\right)}\) | DM1 | 1.1 |
| \([\cos x \approx]\dfrac{4\pi^2 - 16x^2}{4\pi^2 + 4x^2} = \dfrac{\pi^2 - 4x^2}{\pi^2 + x^2}\) | A1 | 1.1 |
## Question 12(a):
$y = \sin\!\left(x + \dfrac{\pi}{2}\right)$ is a translation of $y = \sin x$ | M1 | 3.1a | OR $\sin\!\left(x+\dfrac{\pi}{2}\right) = \sin x\cos\dfrac{\pi}{2} + \sin\dfrac{\pi}{2}\cos x$
$\dfrac{\pi}{2}$ to left and so is the same as $y = \cos x$ | E1 | 2.2a | OR $\sin\!\left(x+\dfrac{\pi}{2}\right) = \sin x \times 0 + 1 \times \cos x = \cos x$. Convincing completion (AG)
**[2 marks]**
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## Question 12(b):
$[\cos x \approx]\dfrac{16\!\left(x+\dfrac{\pi}{2}\right)\!\left(\dfrac{\pi}{2}-x\right)}{5\pi^2 - 4\!\left(x+\dfrac{\pi}{2}\right)\!\left(\dfrac{\pi}{2}-x\right)}$ | M1* | 2.1 | Substitute $x + \dfrac{\pi}{2}$
$[\cos x \approx]\dfrac{16\!\left(\dfrac{\pi^2}{4} - x^2\right)}{5\pi^2 - 4\!\left(\dfrac{\pi^2}{4} - x^2\right)}$ | DM1 | 1.1 | Multiplying out brackets (dep on first M1)
$[\cos x \approx]\dfrac{4\pi^2 - 16x^2}{4\pi^2 + 4x^2} = \dfrac{\pi^2 - 4x^2}{\pi^2 + x^2}$ | A1 | 1.1 | Convincing completion (AG). Check continuation page
**[3 marks]**12
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos x = \sin \left( x + \frac { \pi } { 2 } \right)$.
\item Hence show that $\sin x \approx \frac { 16 x ( \pi - x ) } { 5 \pi ^ { 2 } - 4 x ( \pi - x ) }$ gives the approximation $\cos x \approx \frac { \pi ^ { 2 } - 4 x ^ { 2 } } { \pi ^ { 2 } + x ^ { 2 } }$, as stated in line 31.
\section*{END OF QUESTION PAPER}
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\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q12 [5]}}