| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.8 This is a substantial FP2 question requiring multiple advanced techniques: deriving a multiple angle formula via De Moivre's theorem, solving a quintic by substitution and recognizing sin values, and integrating using the derived identity. While each part follows established FP2 methods, the multi-step nature, the need to connect parts (a) and (b), and the integration requiring careful manipulation of the identity make this significantly harder than average A-level questions but still within standard FP2 territory. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08i Integration by parts4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta\) | B1 | Applies de Moivre correctly |
| Expand using binomial theorem: \(= \cos^5\theta + 5\cos^4\theta(i\sin\theta) + \frac{5\times4}{2!}\cos^3\theta(i\sin\theta)^2 + \frac{5\times4\times3}{3!}\cos^2\theta(i\sin\theta)^3 + \frac{5\times4\times3\times2}{4!}\cos\theta(i\sin\theta)^4 + (i\sin\theta)^5\) | M1 | Uses binomial theorem to expand; may only show imaginary parts |
| \(= \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta\) | A1 | Simplifies coefficients; all imaginary terms correct |
| \(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\) | M1 | Equates imaginary parts; obtains expression for \(\sin 5\theta\) in terms of powers of \(\sin\theta\) |
| \(= 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta\) | ||
| \(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\) | A1 | cso correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(x = \sin\theta\); \(16x^5 - 20x^3 + 5x = -\frac{1}{2} \Rightarrow \sin 5\theta = -\frac{1}{2}\) | M1 | Uses substitution \(x=\sin\theta\); deduces \(\sin 5\theta = \pm\frac{1}{2}\) |
| \(5\theta = 210, 330, 570, 690, 930, 1050, 1290\) (or in radians) or \(210, 570, 930, 1290, 1650\) | A1, A1 | A1 for 3 useable results; A1 for remaining 2 useable results (no repeats in the set of 5) |
| \(\theta = 42, 66, (114), (138), 186, 210, 258\) (or in radians) or \(42, 114, 186, 258, 330\) | dM1 | At least 2 values for \(\theta\) |
| \(\sin\theta = 0.669, 0.914, -0.105, -0.5, -0.978\) | A1 | For the 5 different values of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{\frac{\pi}{4}}(4\sin^5\theta - 5\sin^3\theta)\,d\theta = \frac{1}{4}\int_0^{\frac{\pi}{4}}(\sin 5\theta - 5\sin\theta)\,d\theta\) | M1 | Uses previous work to change the integrand |
| \(= \frac{1}{4}\left[-\frac{1}{5}\cos 5\theta + 5\cos\theta\right]_0^{\frac{\pi}{4}}\) | A1 | Correct result after integrating; limits can be ignored |
| \(\frac{1}{4}\left[-\frac{1}{5}\cos\frac{5\pi}{4} + 5\cos\frac{\pi}{4} - \left(-\frac{1}{5}+5\right)\right]\) | ||
| \(= \frac{1}{4}\left[\frac{1}{5}\times\frac{1}{\sqrt{2}} + \frac{5}{\sqrt{2}} - 4\frac{4}{5}\right]\) | M1 | Substitute given limits; use numerical values for trig functions |
| \(= \dfrac{13\sqrt{2}}{20} - \dfrac{6}{5}\) | A1 | Final answer correct |
| [14] |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$ | B1 | Applies de Moivre correctly |
| Expand using binomial theorem: $= \cos^5\theta + 5\cos^4\theta(i\sin\theta) + \frac{5\times4}{2!}\cos^3\theta(i\sin\theta)^2 + \frac{5\times4\times3}{3!}\cos^2\theta(i\sin\theta)^3 + \frac{5\times4\times3\times2}{4!}\cos\theta(i\sin\theta)^4 + (i\sin\theta)^5$ | M1 | Uses binomial theorem to expand; may only show imaginary parts |
| $= \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$ | A1 | Simplifies coefficients; all imaginary terms correct |
| $\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$ | M1 | Equates imaginary parts; obtains expression for $\sin 5\theta$ in terms of powers of $\sin\theta$ |
| $= 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$ | | |
| $\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ | A1 | cso correct result | (5) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $x = \sin\theta$; $16x^5 - 20x^3 + 5x = -\frac{1}{2} \Rightarrow \sin 5\theta = -\frac{1}{2}$ | M1 | Uses substitution $x=\sin\theta$; deduces $\sin 5\theta = \pm\frac{1}{2}$ |
| $5\theta = 210, 330, 570, 690, 930, 1050, 1290$ (or in radians) or $210, 570, 930, 1290, 1650$ | A1, A1 | A1 for 3 useable results; A1 for remaining 2 useable results (no repeats in the set of 5) |
| $\theta = 42, 66, (114), (138), 186, 210, 258$ (or in radians) or $42, 114, 186, 258, 330$ | dM1 | At least 2 values for $\theta$ |
| $\sin\theta = 0.669, 0.914, -0.105, -0.5, -0.978$ | A1 | For the 5 **different** values of $x$ | (5) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\frac{\pi}{4}}(4\sin^5\theta - 5\sin^3\theta)\,d\theta = \frac{1}{4}\int_0^{\frac{\pi}{4}}(\sin 5\theta - 5\sin\theta)\,d\theta$ | M1 | Uses previous work to change the integrand |
| $= \frac{1}{4}\left[-\frac{1}{5}\cos 5\theta + 5\cos\theta\right]_0^{\frac{\pi}{4}}$ | A1 | Correct result after integrating; limits can be ignored |
| $\frac{1}{4}\left[-\frac{1}{5}\cos\frac{5\pi}{4} + 5\cos\frac{\pi}{4} - \left(-\frac{1}{5}+5\right)\right]$ | | |
| $= \frac{1}{4}\left[\frac{1}{5}\times\frac{1}{\sqrt{2}} + \frac{5}{\sqrt{2}} - 4\frac{4}{5}\right]$ | M1 | Substitute given limits; use numerical values for trig functions |
| $= \dfrac{13\sqrt{2}}{20} - \dfrac{6}{5}$ | A1 | Final answer correct | (4) |
| | | **[14]** |
---7. (a) Use de Moivre's theorem to show that
$$\sin 5 \theta \equiv 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta$$
(b) Hence find the five distinct solutions of the equation
$$16 x ^ { 5 } - 20 x ^ { 3 } + 5 x + \frac { 1 } { 2 } = 0$$
giving your answers to 3 decimal places where necessary.\\
(c) Use the identity given in (a) to find
$$\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 4 \sin ^ { 5 } \theta - 5 \sin ^ { 3 } \theta \right) \mathrm { d } \theta$$
expressing your answer in the form $a \sqrt { } 2 + b$, where $a$ and $b$ are rational numbers.\\
\hfill \mbox{\textit{Edexcel FP2 2014 Q7 [14]}}