| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a structured Further Maths question that guides students through standard De Moivre techniques. Part (a) is algebraic expansion, parts (b)(i-ii) are bookwork results, part (c) follows mechanically from combining previous parts, and part (d) is routine integration. While it requires multiple steps and FM content, the scaffolding makes it accessible and the techniques are standard for F2. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(z+\frac{1}{z}\right)^3\!\left(z-\frac{1}{z}\right)^3 = \left(z^2 - \frac{1}{z^2}\right)^3\) | — | — |
| \(= z^6 - 3z^2 + \frac{3}{z^2} - z^{-6}\) | M1A1 | M1: Attempt to expand; A1: Correct expansion |
| \(= z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right)\) | A1 | Correct answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(z+\frac{1}{z}\right)^3 = z^3 + 3z + \frac{3}{z} + \frac{1}{z^3},\quad \left(z-\frac{1}{z}\right)^3 = z^3 - 3z + \frac{3}{z} - \frac{1}{z^3}\) | M1A1 | M1: Attempt to expand both cubic brackets; A1: Correct expansions |
| \(= z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right)\) | A1 | Correct answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z^n = \cos n\theta + i\sin n\theta\) | B1 | Correct application of de Moivre |
| \(z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \pm\cos n\theta \pm \sin n\theta\) | M1 | Attempt \(z^{-n}\); must be different from their \(z^n\) |
| \(z^n + \frac{1}{z^n} = 2\cos n\theta\)*, \(\quad z^n - \frac{1}{z^n} = 2i\sin n\theta\)* | A1* | \(z^{-n} = \cos n\theta - i\sin n\theta\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(z+\frac{1}{z}\right)^3\!\left(z-\frac{1}{z}\right)^3 = (2\cos\theta)^3(2i\sin\theta)^3\) | B1 | — |
| \(z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right) = 2i\sin 6\theta - 6i\sin 2\theta\) | B1ft | Follow through their \(k\) in place of 3 |
| \(-64i\sin^3\theta\cos^3\theta = 2i\sin 6\theta - 6i\sin 2\theta\) | M1 | Equating right hand sides and simplifying \(2^3 \times (2i)^3\) (B mark needed for each side to gain M mark) |
| \(\cos^3\theta\sin^3\theta = \frac{1}{32}(3\sin 2\theta - \sin 6\theta)\) * | A1cso | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{\pi/8}\cos^3\theta\sin^3\theta\,d\theta = \int_0^{\pi/8}\frac{1}{32}(3\sin 2\theta - \sin 6\theta)\,d\theta\) | — | — |
| \(= \frac{1}{32}\!\left[-\frac{3}{2}\cos 2\theta + \frac{1}{6}\cos 6\theta\right]_0^{\pi/8}\) | M1A1 | M1: \(p\cos 2\theta + q\cos 6\theta\); A1: Correct integration; differentiation scores M0A0 |
| \(= \frac{1}{32}\!\left[\!\left(-\frac{3}{2\sqrt{2}} - \frac{1}{6\sqrt{2}}\right) - \left(-\frac{3}{2} + \frac{1}{6}\right)\right] = \frac{1}{32}\!\left(\frac{4}{3} - \frac{5\sqrt{2}}{6}\right)\) | dM1A1 | dM1: Correct use of limits — lower limit to have non-zero result; dep on previous M mark; A1: Cao (oe) but must be exact |
## Question 8:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^3\!\left(z-\frac{1}{z}\right)^3 = \left(z^2 - \frac{1}{z^2}\right)^3$ | — | — |
| $= z^6 - 3z^2 + \frac{3}{z^2} - z^{-6}$ | M1A1 | M1: Attempt to expand; A1: Correct expansion |
| $= z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right)$ | A1 | Correct answer with no errors seen |
**Alternative:**
| $\left(z+\frac{1}{z}\right)^3 = z^3 + 3z + \frac{3}{z} + \frac{1}{z^3},\quad \left(z-\frac{1}{z}\right)^3 = z^3 - 3z + \frac{3}{z} - \frac{1}{z^3}$ | M1A1 | M1: Attempt to expand both cubic brackets; A1: Correct expansions |
| $= z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right)$ | A1 | Correct answer with no errors |
**Total: (3)**
---
**Part (b)(i)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z^n = \cos n\theta + i\sin n\theta$ | B1 | Correct application of de Moivre |
| $z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \pm\cos n\theta \pm \sin n\theta$ | M1 | Attempt $z^{-n}$; must be different from their $z^n$ |
| $z^n + \frac{1}{z^n} = 2\cos n\theta$*, $\quad z^n - \frac{1}{z^n} = 2i\sin n\theta$* | A1* | $z^{-n} = \cos n\theta - i\sin n\theta$ must be seen |
**Total: (3)**
---
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^3\!\left(z-\frac{1}{z}\right)^3 = (2\cos\theta)^3(2i\sin\theta)^3$ | B1 | — |
| $z^6 - \frac{1}{z^6} - 3\!\left(z^2 - \frac{1}{z^2}\right) = 2i\sin 6\theta - 6i\sin 2\theta$ | B1ft | Follow through their $k$ in place of 3 |
| $-64i\sin^3\theta\cos^3\theta = 2i\sin 6\theta - 6i\sin 2\theta$ | M1 | Equating right hand sides and simplifying $2^3 \times (2i)^3$ (B mark needed for each side to gain M mark) |
| $\cos^3\theta\sin^3\theta = \frac{1}{32}(3\sin 2\theta - \sin 6\theta)$ * | A1cso | — |
**Total: (4)**
---
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\pi/8}\cos^3\theta\sin^3\theta\,d\theta = \int_0^{\pi/8}\frac{1}{32}(3\sin 2\theta - \sin 6\theta)\,d\theta$ | — | — |
| $= \frac{1}{32}\!\left[-\frac{3}{2}\cos 2\theta + \frac{1}{6}\cos 6\theta\right]_0^{\pi/8}$ | M1A1 | M1: $p\cos 2\theta + q\cos 6\theta$; A1: Correct integration; differentiation scores M0A0 |
| $= \frac{1}{32}\!\left[\!\left(-\frac{3}{2\sqrt{2}} - \frac{1}{6\sqrt{2}}\right) - \left(-\frac{3}{2} + \frac{1}{6}\right)\right] = \frac{1}{32}\!\left(\frac{4}{3} - \frac{5\sqrt{2}}{6}\right)$ | dM1A1 | dM1: Correct use of limits — lower limit to have non-zero result; dep on previous M mark; A1: Cao (oe) but must be exact |
**Total: (4)**\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$\left( z + \frac { 1 } { z } \right) ^ { 3 } \left( z - \frac { 1 } { z } \right) ^ { 3 } = z ^ { 6 } - \frac { 1 } { z ^ { 6 } } - k \left( z ^ { 2 } - \frac { 1 } { z ^ { 2 } } \right)$$
where $k$ is a constant to be found.
Given that $z = \cos \theta + \mathrm { i } \sin \theta$, where $\theta$ is real,\\
(b) show that\\
(i) $z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$\\
(ii) $z ^ { n } - \frac { 1 } { z ^ { n } } = 2 \mathrm { i } \sin n \theta$\\
(c) Hence show that
$$\cos ^ { 3 } \theta \sin ^ { 3 } \theta = \frac { 1 } { 32 } ( 3 \sin 2 \theta - \sin 6 \theta )$$
(d) Find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 8 } } \cos ^ { 3 } \theta \sin ^ { 3 } \theta d \theta$$
\hfill \mbox{\textit{Edexcel F2 2015 Q8 [14]}}