| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 This is a classic Further Maths question combining de Moivre's theorem with solving quintic equations. Part (a) is a standard derivation requiring binomial expansion and equating real parts. Part (b) requires the insight to substitute x = cos θ and recognize the connection to part (a), then solve 5θ values—demanding but follows a well-established pattern for this topic. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos5\theta = \text{Re}(\cos\theta + i\sin\theta)^5 = \cos^5\theta - 10\sin^2\theta\cos^3\theta + 5\sin^4\theta\cos\theta\) | M1 A1 | Expands and takes real part. Accept RHS to LHS using \(2\cos\theta = z + \frac{1}{z}\) |
| \(= \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-2\cos^2\theta+\cos^4\theta)\) | M1 | Applies \(\sin^2\theta = 1-\cos^2\theta\) |
| \(= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \cos\theta\), \(\cos5\theta = \frac{1}{2}\sqrt{2}\) | M1 | Applies identity given in (b) |
| \(5\theta = \pm\frac{1}{4}\pi + 2k\pi\) | M1 | Solves \(\cos5\theta = \frac{1}{2}\sqrt{2}\) |
| \(\cos\left(\frac{1}{20}\pi\right)\) | A1 | Gives one correct solution. Accept \(q = \frac{1}{20}\) |
| \(\cos\!\left(\frac{9}{20}\pi\right), \cos\!\left(\frac{17}{20}\pi\right), \cos\!\left(\frac{25}{20}\pi\right), \cos\!\left(\frac{33}{20}\pi\right)\) | A1 | Gives other solutions. OE. A0 for repeated roots |
| Total: 4 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos5\theta = \text{Re}(\cos\theta + i\sin\theta)^5 = \cos^5\theta - 10\sin^2\theta\cos^3\theta + 5\sin^4\theta\cos\theta$ | M1 A1 | Expands and takes real part. Accept RHS to LHS using $2\cos\theta = z + \frac{1}{z}$ |
| $= \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-2\cos^2\theta+\cos^4\theta)$ | M1 | Applies $\sin^2\theta = 1-\cos^2\theta$ |
| $= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ | A1 | AG |
| **Total: 4** | | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \cos\theta$, $\cos5\theta = \frac{1}{2}\sqrt{2}$ | M1 | Applies identity given in (b) |
| $5\theta = \pm\frac{1}{4}\pi + 2k\pi$ | M1 | Solves $\cos5\theta = \frac{1}{2}\sqrt{2}$ |
| $\cos\left(\frac{1}{20}\pi\right)$ | A1 | Gives one correct solution. Accept $q = \frac{1}{20}$ |
| $\cos\!\left(\frac{9}{20}\pi\right), \cos\!\left(\frac{17}{20}\pi\right), \cos\!\left(\frac{25}{20}\pi\right), \cos\!\left(\frac{33}{20}\pi\right)$ | A1 | Gives other solutions. OE. A0 for repeated roots |
| **Total: 4** | | |3
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that
$$\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta$$
\item Hence obtain the roots of the equation
$$32 x ^ { 5 } - 40 x ^ { 3 } + 10 x - \sqrt { 2 } = 0$$
in the form $\cos ( q \pi )$, where $q$ is a rational number.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q3 [8]}}