| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring de Moivre's theorem application, algebraic manipulation to derive a trigonometric identity, connection between polynomial roots and trigonometric values, and use of Vieta's formulas. While the steps are guided, it demands multiple sophisticated techniques and careful algebraic work across three connected parts, placing it well above average difficulty but within reach of strong Further Maths students. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02q De Moivre's theorem: multiple angle formulae4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 5\theta = \text{Im}(c + is)^5 = \text{Im}\left(c^5 + 5c^4is + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5\right)\) | B1 | SOI |
| \(\sin 5\theta = 5c^4s - 10c^2s^3 + s^5\) | M1A1 | |
| \(= s\left(5\left[1-s^2\right]^2 - 10s^2\left[1-s^2\right] + s^4\right)\) | M1 | |
| \(= \ldots = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta\) | A1 | AG |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(\theta = 0, \pm\frac{1}{5}\pi, \pm\frac{2}{5}\pi\) then \(\sin 5\theta = 0\) | B1 | |
| \(\Rightarrow 16s^5 - 20s^3 + 5s = 0\), where \(s = \sin\theta\), \(\Rightarrow s(16s^4 - 20s^2 + 5) = 0\) | B1 | |
| \(s = 0 \Rightarrow \theta = 0\) | B1 | |
| Hence roots of \(16s^4 - 20s^2 + 5 = 0\) are \(\pm\sin\frac{1}{5}\pi, \pm\sin\frac{2}{5}\pi\) | AG | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Since \(\sin\frac{4}{5}\pi = -\sin\left(-\frac{1}{5}\pi\right)\) and \(\sin\frac{3}{5}\pi = -\sin\left(-\frac{2}{5}\pi\right)\) | B1 | |
| \(\sin\left(\frac{4}{5}\pi\right)\sin\left(\frac{3}{5}\pi\right)\sin\left(\frac{2}{5}\pi\right)\sin\left(\frac{1}{5}\pi\right) = \sin\left(-\frac{1}{5}\pi\right)\sin\left(-\frac{2}{5}\pi\right)\sin\left(\frac{1}{5}\pi\right)\sin\left(\frac{2}{5}\pi\right) = \frac{5}{16}\) | M1A1 | |
| \(\sin^2\frac{1}{5}\pi + \sin^2\frac{2}{5}\pi = -\frac{(-20)}{16} = \frac{5}{4}\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) and \(\mathbf{B}\mathbf{e} = \mu\mathbf{e}\) | M1A1 | |
| \(\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}\) | M1 | AG |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda+1)(\lambda^2 - 5\lambda + 6) = 0\) | A1 | |
| \((\lambda+1)(\lambda-2)(\lambda-3) = 0\) | A1 | |
| \(\lambda = -1, 2, 3\) | M1 | |
| Eigenvectors are \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\1\end{pmatrix}\) respectively | A1A1 | Uses either vector product or equations to find eigenvectors |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}3&6&1\\1&-2&-1\\6&6&-2\end{pmatrix}\begin{pmatrix}1\\-1\\0\end{pmatrix} = \begin{pmatrix}-3\\3\\0\end{pmatrix} \Rightarrow \mu_1 = -3\) | M1 | |
| Similarly, other two eigenvalues of \(\mathbf{B}\) are \(-2\) and \(4\) | A1 | |
| Eigenvalues of \(\mathbf{AB}\) are \(3, -4\) and \(12\) | A1 | |
| Corresponding eigenvectors are \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\1\end{pmatrix}\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Closed curve starting and ending at pole, in approximately correct location | B1 | Closed curve starting and ending at pole, in approximately correct location |
| Cardioid with indication of correct scale | B1 | Cardioid with indication of correct scale |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = a(1+\cos\theta) \Rightarrow \sqrt{x^2+y^2} = a\left(1 + \dfrac{x}{\sqrt{x^2+y^2}}\right)\) | M1 | |
| \(x^2 + y^2 = a(x + \sqrt{(x^2+y^2)})\) | A1 | Substitutes for \(r\) and \(\cos(\theta)\) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sector area \(= \dfrac{a^2}{2}\displaystyle\int_0^{\frac{1}{3}\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta\) | M1A1 | |
| \(=\dfrac{a^2}{2}\displaystyle\int_0^{\frac{1}{3}\pi}\left(\dfrac{3}{2}+2\cos\theta+\dfrac{\cos 2\theta}{2}\right)d\theta\) | ||
| \(=\dfrac{a^2}{2}\left[\dfrac{3\theta}{2}+2\sin\theta+\dfrac{\sin 2\theta}{4}\right]_0^{\frac{1}{3}\pi}\) | M1 | |
| \(=\dfrac{a^2}{16}\left(4\pi+9\sqrt{3}\right)\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Arc length \(= \displaystyle\int_0^{\frac{1}{3}\pi}\sqrt{a^2(1+2\cos\theta+\cos^2\theta)+a^2(-\sin\theta)^2}\,d\theta\) | M1A1 | |
| \(= a\displaystyle\int_0^{\frac{1}{3}\pi}\sqrt{2+2\cos\theta}\,d\theta\) | A1 | |
| \(= a\left[4\sin\dfrac{\theta}{2}\right]_0^{\frac{1}{3}\pi} = 2a\) | M1A1 | |
| Total: 5 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 5\theta = \text{Im}(c + is)^5 = \text{Im}\left(c^5 + 5c^4is + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5\right)$ | B1 | SOI |
| $\sin 5\theta = 5c^4s - 10c^2s^3 + s^5$ | M1A1 | |
| $= s\left(5\left[1-s^2\right]^2 - 10s^2\left[1-s^2\right] + s^4\right)$ | M1 | |
| $= \ldots = 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$ | A1 | AG |
| **Total** | **5** | |
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $\theta = 0, \pm\frac{1}{5}\pi, \pm\frac{2}{5}\pi$ then $\sin 5\theta = 0$ | B1 | |
| $\Rightarrow 16s^5 - 20s^3 + 5s = 0$, where $s = \sin\theta$, $\Rightarrow s(16s^4 - 20s^2 + 5) = 0$ | B1 | |
| $s = 0 \Rightarrow \theta = 0$ | B1 | |
| Hence roots of $16s^4 - 20s^2 + 5 = 0$ are $\pm\sin\frac{1}{5}\pi, \pm\sin\frac{2}{5}\pi$ | | AG |
| **Total** | **3** | |
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Since $\sin\frac{4}{5}\pi = -\sin\left(-\frac{1}{5}\pi\right)$ and $\sin\frac{3}{5}\pi = -\sin\left(-\frac{2}{5}\pi\right)$ | B1 | |
| $\sin\left(\frac{4}{5}\pi\right)\sin\left(\frac{3}{5}\pi\right)\sin\left(\frac{2}{5}\pi\right)\sin\left(\frac{1}{5}\pi\right) = \sin\left(-\frac{1}{5}\pi\right)\sin\left(-\frac{2}{5}\pi\right)\sin\left(\frac{1}{5}\pi\right)\sin\left(\frac{2}{5}\pi\right) = \frac{5}{16}$ | M1A1 | |
| $\sin^2\frac{1}{5}\pi + \sin^2\frac{2}{5}\pi = -\frac{(-20)}{16} = \frac{5}{4}$ | A1 | |
| **Total** | **4** | |
## Question 11E(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}\mathbf{e} = \lambda\mathbf{e}$ and $\mathbf{B}\mathbf{e} = \mu\mathbf{e}$ | M1A1 | |
| $\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}$ | M1 | AG |
| **Total** | **3** | |
## Question 11E(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda+1)(\lambda^2 - 5\lambda + 6) = 0$ | A1 | |
| $(\lambda+1)(\lambda-2)(\lambda-3) = 0$ | A1 | |
| $\lambda = -1, 2, 3$ | M1 | |
| Eigenvectors are $\begin{pmatrix}1\\-1\\0\end{pmatrix}$, $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\0\\1\end{pmatrix}$ respectively | A1A1 | Uses either vector product or equations to find eigenvectors |
| **Total** | **6** | |
## Question 11E(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}3&6&1\\1&-2&-1\\6&6&-2\end{pmatrix}\begin{pmatrix}1\\-1\\0\end{pmatrix} = \begin{pmatrix}-3\\3\\0\end{pmatrix} \Rightarrow \mu_1 = -3$ | M1 | |
| Similarly, other two eigenvalues of $\mathbf{B}$ are $-2$ and $4$ | A1 | |
| Eigenvalues of $\mathbf{AB}$ are $3, -4$ and $12$ | A1 | |
| Corresponding eigenvectors are $\begin{pmatrix}1\\-1\\0\end{pmatrix}$, $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\0\\1\end{pmatrix}$ | A1 | |
| **Total** | **4** | |
## Question 11OR(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Closed curve starting and ending at pole, in approximately correct location | B1 | Closed curve starting and ending at pole, in approximately correct location |
| Cardioid with indication of correct scale | B1 | Cardioid with indication of correct scale |
| **Total** | **2** | |
## Question 11OR(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = a(1+\cos\theta) \Rightarrow \sqrt{x^2+y^2} = a\left(1 + \dfrac{x}{\sqrt{x^2+y^2}}\right)$ | M1 | |
| $x^2 + y^2 = a(x + \sqrt{(x^2+y^2)})$ | A1 | Substitutes for $r$ and $\cos(\theta)$ |
| **Total** | **2** | |
## Question 11OR(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sector area $= \dfrac{a^2}{2}\displaystyle\int_0^{\frac{1}{3}\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta$ | M1A1 | |
| $=\dfrac{a^2}{2}\displaystyle\int_0^{\frac{1}{3}\pi}\left(\dfrac{3}{2}+2\cos\theta+\dfrac{\cos 2\theta}{2}\right)d\theta$ | | |
| $=\dfrac{a^2}{2}\left[\dfrac{3\theta}{2}+2\sin\theta+\dfrac{\sin 2\theta}{4}\right]_0^{\frac{1}{3}\pi}$ | M1 | |
| $=\dfrac{a^2}{16}\left(4\pi+9\sqrt{3}\right)$ | A1 | |
| **Total: 4** | | |
---
## Question 11OR(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| Arc length $= \displaystyle\int_0^{\frac{1}{3}\pi}\sqrt{a^2(1+2\cos\theta+\cos^2\theta)+a^2(-\sin\theta)^2}\,d\theta$ | M1A1 | |
| $= a\displaystyle\int_0^{\frac{1}{3}\pi}\sqrt{2+2\cos\theta}\,d\theta$ | A1 | |
| $= a\left[4\sin\dfrac{\theta}{2}\right]_0^{\frac{1}{3}\pi} = 2a$ | M1A1 | |
| **Total: 5** | | |10 (i) Use de Moivre's theorem to show that
$$\sin 5 \theta = 5 \sin \theta - 20 \sin ^ { 3 } \theta + 16 \sin ^ { 5 } \theta$$
(ii) Hence explain why the roots of the equation $16 x ^ { 4 } - 20 x ^ { 2 } + 5 = 0$ are $x = \pm \sin \frac { 1 } { 5 } \pi$ and $x = \pm \sin \frac { 2 } { 5 } \pi$.\\
(iii) Without using a calculator, find the exact values of
$$\sin \frac { 1 } { 5 } \pi \sin \frac { 2 } { 5 } \pi \sin \frac { 3 } { 5 } \pi \sin \frac { 4 } { 5 } \pi \quad \text { and } \quad \sin ^ { 2 } \left( \frac { 1 } { 5 } \pi \right) + \sin ^ { 2 } \left( \frac { 2 } { 5 } \pi \right) .$$
\hfill \mbox{\textit{CAIE FP1 2017 Q10 [12]}}