| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| 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 application, algebraic manipulation to derive a cosecant identity, then reverse-engineering to find roots. The connection between parts (a) and (b) requires insight (substituting x = cosec²θ), and expressing roots in the specific trigonometric form demands careful angle work. However, it follows a well-established pattern for this topic with clear scaffolding between parts. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 7\theta = \text{Im}(c + is)^7\) | M1 | Uses binomial theorem |
| \(\sin 7\theta = -s^7 + 21c^2s^5 - 35c^4s^3 + 7c^6s\) | A1 | |
| \(\sin 7\theta = -s^7 + 21(1-s^2)s^5 - 35(1-s^2)^2s^3 + 7(1-s^2)^3s\) | M1 | Uses \(c^2 = 1 - s^2\) |
| \(\sin 7\theta = -64s^7 + 112s^5 - 56s^3 + 7s\) | A1 | |
| \(\text{cosec}\,7\theta = \frac{1}{-64s^7 + 112s^5 - 56s^3 + 7s}\) | M1 | Divides numerator and denominator by \(s^7\) |
| \(\text{cosec}\,7\theta = \frac{\text{cosec}^7\theta}{7\text{cosec}^6\theta - 56\text{cosec}^4\theta + 112\text{cosec}^2\theta - 64}\) | A1 | AG, CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^7 = 2(7x^6 - 56x^4 + 112x^2 - 64)\) leading to \(\frac{x^7}{7x^6 - 56x^4 + 112x^2 - 64} = 2\) | M1 A1 | Relates with equation in part (a) |
| \(\text{cosec}\,7\theta = 2\) leading to \(\sin 7\theta = \frac{1}{2}\) | M1 | Solves \(\sin 7\theta = \frac{1}{2}\) |
| \(x = \text{cosec}\left(\frac{1}{42}\pi\right)\) | A1 | Gives one correct solution |
| \(x = \text{cosec}(q\pi),\quad q = -\frac{19}{42}, -\frac{11}{42}, -\frac{7}{42}, \frac{5}{42}, \frac{13}{42}, \frac{17}{42}\) | A1 | Gives six other solutions. Allow different values of \(q\) as long as all seven solutions are found |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 7\theta = \text{Im}(c + is)^7$ | M1 | Uses binomial theorem |
| $\sin 7\theta = -s^7 + 21c^2s^5 - 35c^4s^3 + 7c^6s$ | A1 | |
| $\sin 7\theta = -s^7 + 21(1-s^2)s^5 - 35(1-s^2)^2s^3 + 7(1-s^2)^3s$ | M1 | Uses $c^2 = 1 - s^2$ |
| $\sin 7\theta = -64s^7 + 112s^5 - 56s^3 + 7s$ | A1 | |
| $\text{cosec}\,7\theta = \frac{1}{-64s^7 + 112s^5 - 56s^3 + 7s}$ | M1 | Divides numerator and denominator by $s^7$ |
| $\text{cosec}\,7\theta = \frac{\text{cosec}^7\theta}{7\text{cosec}^6\theta - 56\text{cosec}^4\theta + 112\text{cosec}^2\theta - 64}$ | A1 | AG, CWO |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^7 = 2(7x^6 - 56x^4 + 112x^2 - 64)$ leading to $\frac{x^7}{7x^6 - 56x^4 + 112x^2 - 64} = 2$ | M1 A1 | Relates with equation in part (a) |
| $\text{cosec}\,7\theta = 2$ leading to $\sin 7\theta = \frac{1}{2}$ | M1 | Solves $\sin 7\theta = \frac{1}{2}$ |
| $x = \text{cosec}\left(\frac{1}{42}\pi\right)$ | A1 | Gives one correct solution |
| $x = \text{cosec}(q\pi),\quad q = -\frac{19}{42}, -\frac{11}{42}, -\frac{7}{42}, \frac{5}{42}, \frac{13}{42}, \frac{17}{42}$ | A1 | Gives six other solutions. Allow different values of $q$ as long as all seven solutions are found |7
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$\operatorname { cosec } 7 \theta = \frac { \operatorname { cosec } ^ { 7 } \theta } { 7 \operatorname { cosec } ^ { 6 } \theta - 56 \operatorname { cosec } ^ { 4 } \theta + 112 \operatorname { cosec } ^ { 2 } \theta - 64 }$$
\item Hence obtain the roots of the equation
$$x ^ { 7 } - 14 x ^ { 6 } + 112 x ^ { 4 } - 224 x ^ { 2 } + 128 = 0$$
in the form $\operatorname { cosec } q \pi$, where $q$ is rational.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q7 [11]}}