| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.3 This is a standard Further Maths question following a well-established template: derive a multiple angle formula using de Moivre's theorem, then use it to find roots. Part (i) requires careful algebraic manipulation but follows a routine method. Parts (ii) and (iii) involve recognizing that tan(5θ)=0 leads to a polynomial equation and using the sec²θ = 1 + tan²θ identity. While requiring multiple steps and careful algebra, this is a textbook-style question with no novel insights needed, making it moderately above average difficulty. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^24.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta\) | 1 | B1 |
| \((c+is)^5 = c^5 + 5c^4si - 10c^3s^2 - 10ic^2s^3i + 5cs^4 + s^5i\) | 2 | M1A1 |
| \(\tan 5\theta = \dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}\) | 1 | M1 |
| Divide numerator and denominator by \(c^5\): \(\Rightarrow \tan 5\theta = \dfrac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}\) (AG) | 1 | A1 |
| Total | 5 | |
| \(\tan 5\theta = 0 \Rightarrow \theta = \dfrac{1}{5}\pi, \dfrac{2}{5}\pi, \dfrac{3}{5}\pi, \dfrac{4}{5}\pi, \pi\) | 1 | B1 |
| \(t^5 - 10t^3 + 5t = 0\) has roots \(\tan\!\left(\dfrac{1}{5}\pi\right), \tan\!\left(\dfrac{2}{5}\pi\right), \tan\!\left(\dfrac{3}{5}\pi\right), \tan\!\left(\dfrac{4}{5}\pi\right), \tan\pi\) \(\Rightarrow t^4 - 10t^2 + 5 = 0\) has roots \(\tan\!\left(\dfrac{1}{5}\pi\right), \tan\!\left(\dfrac{2}{5}\pi\right), \tan\!\left(\dfrac{3}{5}\pi\right), \tan\!\left(\dfrac{4}{5}\pi\right)\) | 1 | B1 |
| \(\Rightarrow \left(t^2 - \tan^2\!\left(\dfrac{1}{5}\pi\right)\right)\!\left(t^2 - \tan^2\!\left(\dfrac{2}{5}\pi\right)\right) = 0\) since \(\tan\!\left(\dfrac{1}{5}\pi\right) = -\tan\!\left(\dfrac{4}{5}\pi\right)\) and \(\tan\!\left(\dfrac{2}{5}\pi\right) = -\tan\!\left(\dfrac{3}{5}\pi\right)\) | 1 | M1 |
| \(\Rightarrow x^2 - 10x + 5 = 0\) has roots \(\tan^2\!\left(\dfrac{1}{5}\pi\right)\) and \(\tan^2\!\left(\dfrac{2}{5}\pi\right)\) (AG) | 1 | A1 |
| Total | 4 | |
| \(\sec^2\alpha = 1 + \tan^2\alpha\) | 1 | M1 |
| \(y = 1 + x \Rightarrow x = y - 1 \Rightarrow (y-1)^2 - 10(y-1) + 5 = 0\) | 1 | M1 |
| \(\Rightarrow y^2 - 12y + 16 = 0\) | 1 | A1 |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 0 \\ -1 & 0 & 4 \end{vmatrix} = \begin{pmatrix}8\\4\\2\end{pmatrix} \sim \begin{pmatrix}4\\2\\1\end{pmatrix}\) | 2 | M1A1 |
| \(\dfrac{\begin{pmatrix}1\\0\\0\end{pmatrix}\cdot\begin{pmatrix}4\\2\\1\end{pmatrix}}{\sqrt{4^2+2^2+1^2}} = \dfrac{4}{\sqrt{21}}\) (AG) | 2 | M1A1 |
| Total | 4 | |
| \(\mathbf{p} = \dfrac{3}{\sqrt{21}}\left(\dfrac{4\mathbf{i}+2\mathbf{j}+\mathbf{k}}{\sqrt{21}}\right) = \dfrac{1}{7}(4\mathbf{i}+2\mathbf{j}+\mathbf{k})\) | 1 | B1 |
| Line \(AP\): \(\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + t\begin{pmatrix}-3\\2\\1\end{pmatrix}\) | 2 | M1A1 |
| For \(Q\): \(1 - 3t = 0 \Rightarrow t = \dfrac{1}{3} \Rightarrow \mathbf{q} = \dfrac{1}{3}\begin{pmatrix}0\\2\\1\end{pmatrix}\) | 2 | M1A1 |
| Total | 5 | |
| \(\overrightarrow{AB} = \begin{pmatrix}-1\\2\\0\end{pmatrix},\ \overrightarrow{BQ} = \dfrac{1}{3}\begin{pmatrix}0\\-4\\1\end{pmatrix}\) | 1 | B1 |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&2&0\\0&-4&1\end{vmatrix} = \begin{pmatrix}2\\1\\4\end{pmatrix}\) | 2 | M1A1 |
| \(\cos^{-1}\left(\dfrac{\begin{pmatrix}4\\2\\1\end{pmatrix}\cdot\begin{pmatrix}2\\1\\4\end{pmatrix}}{\sqrt{21}\cdot\sqrt{21}}\right) = \cos^{-1}\dfrac{8+2+4}{21} = \cos^{-1}\dfrac{14}{21} = \cos^{-1}\dfrac{2}{3}\) (AG) | 2 | M1A1 |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Closed curve through pole with correct orientation. Completely correct. | 2 | B1, B1 |
| \(2 \times \dfrac{1}{2}a^2\displaystyle\int_{\frac{1}{2}\pi}^{\pi}(1-2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta = a^2\int_{\frac{1}{2}\pi}^{\pi}\!\left(\dfrac{3}{2}-2\cos\theta+\dfrac{1}{2}\cos 2\theta\right)\mathrm{d}\theta\) | 2 | M1M1 |
| \(= a^2\!\left[\dfrac{3}{2}\theta - 2\sin\theta + \dfrac{1}{4}\sin 2\theta\right]_{\frac{1}{2}\pi}^{\pi}\) | 2 | M1A1 |
| \(= a^2\!\left(\dfrac{3}{4}\pi + 2\right)\) | 1 | A1 |
| Total | 5 | |
| \(\left(\dfrac{\mathrm{d}s}{\mathrm{d}\theta}\right)^2 = a^2(1 - 2\cos\theta + \cos^2\theta + \sin^2\theta)\) | 1 | B1 |
| \(= 2a^2(1-\cos\theta) = 2a^2 \cdot 2\sin^2\dfrac{1}{2}\theta = 4a^2\sin^2\dfrac{1}{2}\theta\) (AG) | 2 | M1A1 |
| \(s = 2\times\displaystyle\int_{\frac{1}{2}\pi}^{\pi} 2a\sin\dfrac{1}{2}\theta\,\mathrm{d}\theta\) | 1 | M1 |
| \(= 4a\!\left[-2\cos\dfrac{1}{2}\theta\right]_{\frac{1}{2}\pi}^{\pi}\) | 1 | A1 |
| \(= 4\sqrt{2}\,a\) | 2 | M1A1 |
| Total | 7 |
## Question 10:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$ | 1 | B1 | |
| $(c+is)^5 = c^5 + 5c^4si - 10c^3s^2 - 10ic^2s^3i + 5cs^4 + s^5i$ | 2 | M1A1 | |
| $\tan 5\theta = \dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}$ | 1 | M1 | |
| Divide numerator and denominator by $c^5$: $\Rightarrow \tan 5\theta = \dfrac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}$ (AG) | 1 | A1 | |
| **Total** | **5** | | |
| $\tan 5\theta = 0 \Rightarrow \theta = \dfrac{1}{5}\pi, \dfrac{2}{5}\pi, \dfrac{3}{5}\pi, \dfrac{4}{5}\pi, \pi$ | 1 | B1 | |
| $t^5 - 10t^3 + 5t = 0$ has roots $\tan\!\left(\dfrac{1}{5}\pi\right), \tan\!\left(\dfrac{2}{5}\pi\right), \tan\!\left(\dfrac{3}{5}\pi\right), \tan\!\left(\dfrac{4}{5}\pi\right), \tan\pi$ $\Rightarrow t^4 - 10t^2 + 5 = 0$ has roots $\tan\!\left(\dfrac{1}{5}\pi\right), \tan\!\left(\dfrac{2}{5}\pi\right), \tan\!\left(\dfrac{3}{5}\pi\right), \tan\!\left(\dfrac{4}{5}\pi\right)$ | 1 | B1 | |
| $\Rightarrow \left(t^2 - \tan^2\!\left(\dfrac{1}{5}\pi\right)\right)\!\left(t^2 - \tan^2\!\left(\dfrac{2}{5}\pi\right)\right) = 0$ since $\tan\!\left(\dfrac{1}{5}\pi\right) = -\tan\!\left(\dfrac{4}{5}\pi\right)$ and $\tan\!\left(\dfrac{2}{5}\pi\right) = -\tan\!\left(\dfrac{3}{5}\pi\right)$ | 1 | M1 | |
| $\Rightarrow x^2 - 10x + 5 = 0$ has roots $\tan^2\!\left(\dfrac{1}{5}\pi\right)$ and $\tan^2\!\left(\dfrac{2}{5}\pi\right)$ (AG) | 1 | A1 | |
| **Total** | **4** | | |
| $\sec^2\alpha = 1 + \tan^2\alpha$ | 1 | M1 | |
| $y = 1 + x \Rightarrow x = y - 1 \Rightarrow (y-1)^2 - 10(y-1) + 5 = 0$ | 1 | M1 | |
| $\Rightarrow y^2 - 12y + 16 = 0$ | 1 | A1 | |
| **Total** | **3** | | |
---
## Question 11E:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 0 \\ -1 & 0 & 4 \end{vmatrix} = \begin{pmatrix}8\\4\\2\end{pmatrix} \sim \begin{pmatrix}4\\2\\1\end{pmatrix}$ | 2 | M1A1 | |
| $\dfrac{\begin{pmatrix}1\\0\\0\end{pmatrix}\cdot\begin{pmatrix}4\\2\\1\end{pmatrix}}{\sqrt{4^2+2^2+1^2}} = \dfrac{4}{\sqrt{21}}$ (AG) | 2 | M1A1 | |
| **Total** | **4** | | |
| $\mathbf{p} = \dfrac{3}{\sqrt{21}}\left(\dfrac{4\mathbf{i}+2\mathbf{j}+\mathbf{k}}{\sqrt{21}}\right) = \dfrac{1}{7}(4\mathbf{i}+2\mathbf{j}+\mathbf{k})$ | 1 | B1 | |
| Line $AP$: $\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + t\begin{pmatrix}-3\\2\\1\end{pmatrix}$ | 2 | M1A1 | |
| For $Q$: $1 - 3t = 0 \Rightarrow t = \dfrac{1}{3} \Rightarrow \mathbf{q} = \dfrac{1}{3}\begin{pmatrix}0\\2\\1\end{pmatrix}$ | 2 | M1A1 | |
| **Total** | **5** | | |
| $\overrightarrow{AB} = \begin{pmatrix}-1\\2\\0\end{pmatrix},\ \overrightarrow{BQ} = \dfrac{1}{3}\begin{pmatrix}0\\-4\\1\end{pmatrix}$ | 1 | B1 | |
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&2&0\\0&-4&1\end{vmatrix} = \begin{pmatrix}2\\1\\4\end{pmatrix}$ | 2 | M1A1 | |
| $\cos^{-1}\left(\dfrac{\begin{pmatrix}4\\2\\1\end{pmatrix}\cdot\begin{pmatrix}2\\1\\4\end{pmatrix}}{\sqrt{21}\cdot\sqrt{21}}\right) = \cos^{-1}\dfrac{8+2+4}{21} = \cos^{-1}\dfrac{14}{21} = \cos^{-1}\dfrac{2}{3}$ (AG) | 2 | M1A1 | |
| **Total** | **5** | | |
---
## Question 11O:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Closed curve through pole with correct orientation. Completely correct. | 2 | B1, B1 | |
| $2 \times \dfrac{1}{2}a^2\displaystyle\int_{\frac{1}{2}\pi}^{\pi}(1-2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta = a^2\int_{\frac{1}{2}\pi}^{\pi}\!\left(\dfrac{3}{2}-2\cos\theta+\dfrac{1}{2}\cos 2\theta\right)\mathrm{d}\theta$ | 2 | M1M1 | |
| $= a^2\!\left[\dfrac{3}{2}\theta - 2\sin\theta + \dfrac{1}{4}\sin 2\theta\right]_{\frac{1}{2}\pi}^{\pi}$ | 2 | M1A1 | |
| $= a^2\!\left(\dfrac{3}{4}\pi + 2\right)$ | 1 | A1 | |
| **Total** | **5** | | |
| $\left(\dfrac{\mathrm{d}s}{\mathrm{d}\theta}\right)^2 = a^2(1 - 2\cos\theta + \cos^2\theta + \sin^2\theta)$ | 1 | B1 | |
| $= 2a^2(1-\cos\theta) = 2a^2 \cdot 2\sin^2\dfrac{1}{2}\theta = 4a^2\sin^2\dfrac{1}{2}\theta$ (AG) | 2 | M1A1 | |
| $s = 2\times\displaystyle\int_{\frac{1}{2}\pi}^{\pi} 2a\sin\dfrac{1}{2}\theta\,\mathrm{d}\theta$ | 1 | M1 | |
| $= 4a\!\left[-2\cos\dfrac{1}{2}\theta\right]_{\frac{1}{2}\pi}^{\pi}$ | 1 | A1 | |
| $= 4\sqrt{2}\,a$ | 2 | M1A1 | |
| **Total** | **7** | | |10 (i) Using de Moivre's theorem, show that
$$\tan 5 \theta = \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta } .$$
(ii) Hence show that the equation $x ^ { 2 } - 10 x + 5 = 0$ has roots $\tan ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)$ and $\tan ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)$.\\
(iii) Deduce a quadratic equation, with integer coefficients, having roots $\sec ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)$ and $\sec ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)$. [3]\\
\hfill \mbox{\textit{CAIE FP1 2017 Q10 [12]}}