| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question requiring De Moivre's theorem and binomial expansion to express cos^6(θ) as a sum of multiple angles, followed by routine integration. While it requires multiple steps and is harder than typical A-level content, it follows a well-established template taught in FP3 with no novel insight needed. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.08a Maclaurin series: find series for function4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(z^n = \cos n\theta + i \sin n\theta\) | B1 | For applying de Moivre's theorem |
| \(z^{-n} = \cos n\theta - i \sin n\theta\), hence \(z^n + z^{-n} = 2\cos n\theta\) | B1 | For complete proof |
| 2 | ||
| (ii) \(2^6 \cos^6 \theta = (z + z^{-1})^6\) | M1 | For considering \((z + z^{-1})^6\) |
| \(= (z^6 + z^{-6}) + 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) + 20\) | M1 | For expanding and grouping terms |
| \(= 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\) | A1 | For correct substitution of multiple angles |
| Hence \(\cos^6 \theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\) | A1 | For correct answer |
| Integral is \(\frac{1}{32}\left[\frac{1}{6}\sin 6\theta + \frac{3}{2}\sin 4\theta + \frac{15}{2}\sin 2\theta + 10\theta\right]_0^{\frac{\pi}{2}}\) | M1 | For integrating multiple angle expression |
| A1√ | For correct terms | |
| \(= \frac{1}{32}\left[\frac{1}{6} \times 0 + \frac{3}{2} \times (-\frac{1}{\sqrt{3}}) + \frac{15}{2} \times (\frac{1}{\sqrt{3}}) + 10 \times \frac{\pi}{4}\right]\) | M1 | For use of limits |
| A1 | For correct answer | |
| 8 | ||
| 10 |
**(i)** $z^n = \cos n\theta + i \sin n\theta$ | B1 | For applying de Moivre's theorem
$z^{-n} = \cos n\theta - i \sin n\theta$, hence $z^n + z^{-n} = 2\cos n\theta$ | B1 | For complete proof
| **2** |
**(ii)** $2^6 \cos^6 \theta = (z + z^{-1})^6$ | M1 | For considering $(z + z^{-1})^6$
$= (z^6 + z^{-6}) + 6(z^4 + z^{-4}) + 15(z^2 + z^{-2}) + 20$ | M1 | For expanding and grouping terms
$= 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20$ | A1 | For correct substitution of multiple angles
Hence $\cos^6 \theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$ | A1 | For correct answer
Integral is $\frac{1}{32}\left[\frac{1}{6}\sin 6\theta + \frac{3}{2}\sin 4\theta + \frac{15}{2}\sin 2\theta + 10\theta\right]_0^{\frac{\pi}{2}}$ | M1 | For integrating multiple angle expression
| A1√ | For correct terms
$= \frac{1}{32}\left[\frac{1}{6} \times 0 + \frac{3}{2} \times (-\frac{1}{\sqrt{3}}) + \frac{15}{2} \times (\frac{1}{\sqrt{3}}) + 10 \times \frac{\pi}{4}\right]$ | M1 | For use of limits
| A1 | For correct answer
| **8** |
| **10** |7 (i) Prove that if $z = \mathrm { e } ^ { \mathrm { i } \theta }$, then $z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$.\\
(ii) Express $\cos ^ { 6 } \theta$ in terms of cosines of multiples of $\theta$, and hence find the exact value of
$$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \cos ^ { 6 } \theta \mathrm {~d} \theta$$
\hfill \mbox{\textit{OCR FP3 Q7 [10]}}