| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.3 This is a standard Further Maths FP3 question testing De Moivre's theorem and binomial expansion in a structured, multi-part format. Part (i) requires routine application of exponential forms and binomial expansion, part (ii) is a direct substitution using a standard identity, and part (iii) involves straightforward integration of trigonometric functions. While it requires multiple techniques and careful algebraic manipulation, the question provides clear scaffolding and follows a predictable pattern typical of FP3 exam questions, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08d Evaluate definite integrals: between limits4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin\theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})\) | B1 | \(z\) may be used for \(e^{i\theta}\) throughout. For expression for \(\sin\theta\) seen or implied |
| Expanding \((e^{i\theta} - e^{-i\theta})^6\) | M1 | For expanding \((e^{i\theta} - e^{-i\theta})^6\). At least 4 terms and 3 binomial coefficients required |
| \(\sin^6\theta = -\frac{1}{64}(e^{6i\theta} - 6e^{4i\theta} + 15e^{2i\theta} - 20 + 15e^{-2i\theta} - 6e^{-4i\theta} + e^{-6i\theta})\) | A1 | For correct expansion. Allow \(\frac{\pm(i)}{64}(\ldots)\) |
| \(= -\frac{1}{64}(2\cos 6\theta - 12\cos 4\theta + 30\cos 2\theta - 20)\) | M1 | For grouping terms and using multiple angles |
| \(\sin^6\theta = -\frac{1}{32}(\cos 6\theta - 6\cos 4\theta + 15\cos 2\theta - 10)\) | A1 5 | For answer obtained correctly AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos^6\theta = OR\; \sin^6\!\left(\tfrac{1}{2}\pi - \theta\right) = -\frac{1}{32}(\cos(3\pi - 6\theta) - 6\cos(2\pi - 4\theta) + 15\cos(\pi - 2\theta) - 10)\) | M1 | For substituting \(\left(\frac{1}{2}\pi - \theta\right)\) for \(\theta\) throughout |
| A1 | For correct unsimplified expression | |
| \(\cos^6\theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\) | A1 3 | For correct expression with \(\cos n\theta\) terms AEF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^{\frac{1}{4}\pi} \frac{1}{32}(-2\cos 6\theta - 30\cos 2\theta)\,d\theta\) | B1\(\sqrt{}\) | For correct integral, f.t. from \(\sin^6\theta - \cos^6\theta\) |
| \(= -\frac{1}{16}\left[\frac{1}{6}\sin 6\theta + \frac{15}{2}\sin 2\theta\right]_0^{\frac{1}{4}\pi}\) | M1, A1\(\sqrt{}\) | For integrating \(\cos n\theta\), \(\sin n\theta\) or \(e^{in\theta}\); For correct integration, f.t. from integrand |
| \(= -\frac{11}{24}\) | A1 4 | For correct answer WWW |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin\theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})$ | B1 | $z$ may be used for $e^{i\theta}$ throughout. For expression for $\sin\theta$ seen or implied |
| Expanding $(e^{i\theta} - e^{-i\theta})^6$ | M1 | For expanding $(e^{i\theta} - e^{-i\theta})^6$. At least 4 terms and 3 binomial coefficients required |
| $\sin^6\theta = -\frac{1}{64}(e^{6i\theta} - 6e^{4i\theta} + 15e^{2i\theta} - 20 + 15e^{-2i\theta} - 6e^{-4i\theta} + e^{-6i\theta})$ | A1 | For correct expansion. Allow $\frac{\pm(i)}{64}(\ldots)$ |
| $= -\frac{1}{64}(2\cos 6\theta - 12\cos 4\theta + 30\cos 2\theta - 20)$ | M1 | For grouping terms and using multiple angles |
| $\sin^6\theta = -\frac{1}{32}(\cos 6\theta - 6\cos 4\theta + 15\cos 2\theta - 10)$ | A1 **5** | For answer obtained correctly **AG** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos^6\theta = OR\; \sin^6\!\left(\tfrac{1}{2}\pi - \theta\right) = -\frac{1}{32}(\cos(3\pi - 6\theta) - 6\cos(2\pi - 4\theta) + 15\cos(\pi - 2\theta) - 10)$ | M1 | For substituting $\left(\frac{1}{2}\pi - \theta\right)$ for $\theta$ throughout |
| | A1 | For correct unsimplified expression |
| $\cos^6\theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$ | A1 **3** | For correct expression with $\cos n\theta$ terms **AEF** |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{1}{4}\pi} \frac{1}{32}(-2\cos 6\theta - 30\cos 2\theta)\,d\theta$ | B1$\sqrt{}$ | For correct integral, f.t. from $\sin^6\theta - \cos^6\theta$ |
| $= -\frac{1}{16}\left[\frac{1}{6}\sin 6\theta + \frac{15}{2}\sin 2\theta\right]_0^{\frac{1}{4}\pi}$ | M1, A1$\sqrt{}$ | For integrating $\cos n\theta$, $\sin n\theta$ or $e^{in\theta}$; For correct integration, f.t. from integrand |
| $= -\frac{11}{24}$ | A1 **4** | For correct answer **WWW** |
**Total: 12 marks**8 (i) By expressing $\sin \theta$ in terms of $\mathrm { e } ^ { \mathrm { i } \theta }$ and $\mathrm { e } ^ { - \mathrm { i } \theta }$, show that
$$\sin ^ { 6 } \theta \equiv - \frac { 1 } { 32 } ( \cos 6 \theta - 6 \cos 4 \theta + 15 \cos 2 \theta - 10 )$$
(ii) Replace $\theta$ by ( $\frac { 1 } { 2 } \pi - \theta$ ) in the identity in part (i) to obtain a similar identity for $\cos ^ { 6 } \theta$.\\
(iii) Hence find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \sin ^ { 6 } \theta - \cos ^ { 6 } \theta \right) \mathrm { d } \theta$.
\hfill \mbox{\textit{OCR FP3 2009 Q8 [12]}}