| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring de Moivre's theorem manipulation to derive a cosec identity, then cleverly connecting it to polynomial roots. Part (a) involves substantial algebraic manipulation with multiple substitutions, while part (b) requires recognizing the connection between the derived identity and the polynomial structure. The multi-step reasoning and non-standard application elevate this above typical Further Maths exercises, though the path is relatively guided once the connection is spotted. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cos 5\theta + i\sin 5\theta) = (c + is)^5\) | M1 | Uses binomial theorem |
| \(\sin 5\theta = s^5 - 10c^2s^3 + 5c^4s\) | A1 | |
| \(s^5 - 10(1-s^2)s^3 + 5(1-s^2)^2 s\) | M1 | Uses \(c^2 = 1 - s^2\) or \(\cot^2\theta = \csc^2\theta - 1\) after dividing numerator and denominator by \(s^5\) |
| \(16s^5 - 20s^3 + 5s\) | A1 | |
| \(\csc 5\theta = \frac{1}{16s^5 - 20s^3 + 5s} \times \frac{\csc^5\theta}{\csc^5\theta}\) | M1 | Divides simplified numerator and denominator by \(s^5\) |
| \(\frac{\csc^5\theta}{5\csc^4\theta - 20\csc^2\theta + 16}\) | A1 | AG |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^5 = 2(5x^4 - 20x^2 + 16)\) leading to \(\frac{x^5}{5x^4 - 20x^2 + 16} = 2\) | M1 | Relates with equation in part (a) |
| \(\csc 5\theta = 2\) leading to \(\sin 5\theta = \frac{1}{2}\) | M1 | Solves \(\sin 5\theta = \frac{1}{2}\) |
| \(x = \csc\!\left(\frac{1}{30}\pi\right)\) | A1 | Gives one correct solution |
| \(\csc\!\left(\frac{5}{30}\pi\right), \csc\!\left(-\frac{7}{30}\pi\right), \csc\!\left(-\frac{11}{30}\pi\right), \csc\!\left(\frac{13}{30}\pi\right)\) | A1 | Gives four other solutions. Allow different values of \(q\) as long as all five solutions are found. \(9.56, 2, -1.49, -1.09, 1.02\) |
| 4 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cos 5\theta + i\sin 5\theta) = (c + is)^5$ | M1 | Uses binomial theorem |
| $\sin 5\theta = s^5 - 10c^2s^3 + 5c^4s$ | A1 | |
| $s^5 - 10(1-s^2)s^3 + 5(1-s^2)^2 s$ | M1 | Uses $c^2 = 1 - s^2$ or $\cot^2\theta = \csc^2\theta - 1$ after dividing numerator and denominator by $s^5$ |
| $16s^5 - 20s^3 + 5s$ | A1 | |
| $\csc 5\theta = \frac{1}{16s^5 - 20s^3 + 5s} \times \frac{\csc^5\theta}{\csc^5\theta}$ | M1 | Divides simplified numerator and denominator by $s^5$ |
| $\frac{\csc^5\theta}{5\csc^4\theta - 20\csc^2\theta + 16}$ | A1 | AG |
| | **6** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^5 = 2(5x^4 - 20x^2 + 16)$ leading to $\frac{x^5}{5x^4 - 20x^2 + 16} = 2$ | M1 | Relates with equation in part (a) |
| $\csc 5\theta = 2$ leading to $\sin 5\theta = \frac{1}{2}$ | M1 | Solves $\sin 5\theta = \frac{1}{2}$ |
| $x = \csc\!\left(\frac{1}{30}\pi\right)$ | A1 | Gives one correct solution |
| $\csc\!\left(\frac{5}{30}\pi\right), \csc\!\left(-\frac{7}{30}\pi\right), \csc\!\left(-\frac{11}{30}\pi\right), \csc\!\left(\frac{13}{30}\pi\right)$ | A1 | Gives four other solutions. Allow different values of $q$ as long as all five solutions are found. $9.56, 2, -1.49, -1.09, 1.02$ |
| | **4** | |
---6
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$\operatorname { cosec } 5 \theta = \frac { \operatorname { cosec } ^ { 5 } \theta } { 5 \operatorname { cosec } ^ { 4 } \theta - 20 \operatorname { cosec } ^ { 2 } \theta + 16 }$$
\item Hence obtain the roots of the equation
$$x ^ { 5 } - 10 x ^ { 4 } + 40 x ^ { 2 } - 32 = 0$$
in the form $\operatorname { cosec } ( q \pi )$, where $q$ is rational.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q6 [10]}}