| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Standard +0.3 This is a standard FP3 question on de Moivre's theorem and multiple angle formulas. Part (i) requires routine application of Euler's formula and binomial expansion to derive a known identity. Part (ii) is straightforward integration of the result. While it involves complex numbers and multiple steps, it follows a well-practiced technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})\) | B1 | \(z\) or \(e^{i\theta}\) may be used throughout. For correct expression for \(\sin\theta\) soi |
| \(\sin^4\theta = \frac{1}{16}(z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4})\) | M1 | For expanding \((e^{i\theta} - e^{-i\theta})^4\) (with at least 3 terms and 1 binomial coefficient) |
| \(\Rightarrow \sin^4\theta = \frac{1}{16}(2\cos 4\theta - 8\cos 2\theta + 6)\) | M1 | For grouping terms and using multiple angles |
| \(\Rightarrow \sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)\) | A1 4 | For answer obtained correctly AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{1}{8}[\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta]_0^{\frac{\pi}{2}}\) | M1 | For integrating \((0 \text{ to } 4\sin 4\theta + 8\sin 2\theta + C\theta)\) |
| \(= \frac{1}{8}(\frac{1}{8}\sqrt{3} - \sqrt{3} + \frac{1}{8}\pi) = \frac{1}{64}(4\pi - 7\sqrt{3})\) | M1 | For completing integration and substituting limits |
| A1 4 | For correct answer AEF(exact) |
## (i)
$\sin\theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})$ | B1 | $z$ or $e^{i\theta}$ may be used throughout. For correct expression for $\sin\theta$ soi
$\sin^4\theta = \frac{1}{16}(z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4})$ | M1 | For expanding $(e^{i\theta} - e^{-i\theta})^4$ (with at least 3 terms and 1 binomial coefficient)
$\Rightarrow \sin^4\theta = \frac{1}{16}(2\cos 4\theta - 8\cos 2\theta + 6)$ | M1 | For grouping terms and using multiple angles
$\Rightarrow \sin^4\theta = \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)$ | A1 4 | For answer obtained correctly **AG**
## (ii)
$\int_0^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{1}{8}[\frac{1}{4}\sin 4\theta - 2\sin 2\theta + 3\theta]_0^{\frac{\pi}{2}}$ | M1 | For integrating $(0 \text{ to } 4\sin 4\theta + 8\sin 2\theta + C\theta)$
$= \frac{1}{8}(\frac{1}{8}\sqrt{3} - \sqrt{3} + \frac{1}{8}\pi) = \frac{1}{64}(4\pi - 7\sqrt{3})$ | M1 | For completing integration and substituting limits
| A1 4 | For correct answer **AEF(exact)**
---\begin{enumerate}[label=(\roman*)]
\item Express $\sin \theta$ in terms of $e^{i\theta}$ and $e^{-i\theta}$ and show that
$$\sin^4 \theta \equiv \frac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3).$$ [4]
\item Hence find the exact value of $\int_0^{\frac{\pi}{4}} \sin^4 \theta \, d\theta$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q3 [8]}}