| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP3 question following a well-established template: derive a multiple angle formula using de Moivre's theorem, then use it to solve a polynomial by recognizing the connection to Chebyshev polynomials. While it requires multiple steps and knowledge of Further Maths content, the path is algorithmic and this type appears regularly in past papers, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5\) | B1 | Or \(\cos 5\theta = \text{re}\{(\cos\theta+i\sin\theta)^5\}\) |
| \(= c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5\) | M1 | No more than 1 error, can be unsimplified |
| \(\cos 5\theta = c^5 - 10c^3s^2 + 5cs^4\) | M1 | Take real parts |
| \(= c^5 - 10c^3(1-c^2) + 5c(1-c^2)^2\) | M1 | |
| \(= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5\) | ||
| \(\cos 5\theta = 16c^5 - 20c^3 + 5c\) | A1 | AG |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Multiplying by \(x\) gives \(16x^5 - 20x^3 + 5x = 0\) | Hence, so no marks for using quadratic at this stage. | |
| letting \(x=\cos\alpha\) gives \(\cos 5\alpha = 0\) | M1 | |
| hence \(5\alpha = \frac{1}{2}\pi, \frac{3}{2}\pi, \frac{5}{2}\pi, \frac{7}{2}\pi, \frac{9}{2}\pi\) | A1 | |
| \(\alpha = \frac{1}{10}\pi, \frac{3}{10}\pi, \frac{5}{10}\pi, \frac{7}{10}\pi, \frac{9}{10}\pi\) | ||
| \(\cos\frac{5}{10}\pi = 0\) which is not a root | A1 | |
| so roots \(x = \cos\frac{1}{10}\pi,\ \cos\frac{3}{10}\pi,\ \cos\frac{7}{10}\pi,\ \cos\frac{9}{10}\pi\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16x^4 - 20x^2 + 5 = 0 \Leftrightarrow x^2 = \dfrac{20\pm\sqrt{80}}{32}\) | B1 | Can be gained if seen in (ii) |
| cos decreases between 0 and \(\pi\) so \(\cos\frac{1}{10}\pi\) is greatest root | M1 | |
| so \(\cos\frac{1}{10}\pi = \sqrt{\dfrac{20+\sqrt{80}}{32}} = \sqrt{\dfrac{5+\sqrt{5}}{8}}\) | A1 | Dep on full marks in (ii) |
| [3] |
# Question 8:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5$ | B1 | Or $\cos 5\theta = \text{re}\{(\cos\theta+i\sin\theta)^5\}$ |
| $= c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5$ | M1 | No more than 1 error, can be unsimplified |
| $\cos 5\theta = c^5 - 10c^3s^2 + 5cs^4$ | M1 | Take real parts |
| $= c^5 - 10c^3(1-c^2) + 5c(1-c^2)^2$ | M1 | |
| $= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5$ | | |
| $\cos 5\theta = 16c^5 - 20c^3 + 5c$ | A1 | AG |
| | **[5]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Multiplying by $x$ gives $16x^5 - 20x^3 + 5x = 0$ | | Hence, so no marks for using quadratic at this stage. |
| letting $x=\cos\alpha$ gives $\cos 5\alpha = 0$ | M1 | |
| hence $5\alpha = \frac{1}{2}\pi, \frac{3}{2}\pi, \frac{5}{2}\pi, \frac{7}{2}\pi, \frac{9}{2}\pi$ | A1 | |
| $\alpha = \frac{1}{10}\pi, \frac{3}{10}\pi, \frac{5}{10}\pi, \frac{7}{10}\pi, \frac{9}{10}\pi$ | | |
| $\cos\frac{5}{10}\pi = 0$ which is not a root | A1 | |
| so roots $x = \cos\frac{1}{10}\pi,\ \cos\frac{3}{10}\pi,\ \cos\frac{7}{10}\pi,\ \cos\frac{9}{10}\pi$ | A1 | |
| | **[4]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16x^4 - 20x^2 + 5 = 0 \Leftrightarrow x^2 = \dfrac{20\pm\sqrt{80}}{32}$ | B1 | Can be gained if seen in (ii) |
| cos decreases between 0 and $\pi$ so $\cos\frac{1}{10}\pi$ is greatest root | M1 | |
| so $\cos\frac{1}{10}\pi = \sqrt{\dfrac{20+\sqrt{80}}{32}} = \sqrt{\dfrac{5+\sqrt{5}}{8}}$ | A1 | Dep on full marks in (ii) |
| | **[3]** | |8 (i) Use de Moivre's theorem to show that $\cos 5 \theta \equiv 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta$.\\
(ii) Hence find the roots of $16 x ^ { 4 } - 20 x ^ { 2 } + 5 = 0$ in the form $\cos \alpha$ where $0 \leqslant \alpha \leqslant \pi$.\\
(iii) Hence find the exact value of $\cos \frac { 1 } { 10 } \pi$.
\hfill \mbox{\textit{OCR FP3 2013 Q8 [12]}}