| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Solve equations using trigonometric identities |
| Difficulty | Challenging +1.3 This is a Further Maths question combining de Moivre's theorem with trigonometric identities and equation solving. Part (a) is a standard application of binomial expansion and de Moivre's theorem to derive a triple angle formula—routine for Further Maths students. Part (b) requires algebraic manipulation of the derived identity to solve an equation, which is moderately challenging but follows a clear path once the substitution is made. The question is harder than typical A-level maths but represents standard Further Maths content without requiring exceptional insight. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}\right)^3\) | M1 | Unsimplified. Allow cis notation |
| \(= \cos^3\frac{\theta}{3} + 3\cos^2\frac{\theta}{3}\left(i\sin\frac{\theta}{3}\right) + 3\cos\frac{\theta}{3}\left(i\sin\frac{\theta}{3}\right)^2 + \left(i\sin\frac{\theta}{3}\right)^3\) | A1 | |
| \(= \cos^3\frac{\theta}{3} + 3i\cos^2\frac{\theta}{3}\sin\frac{\theta}{3} - 3\cos\frac{\theta}{3}\sin^2\frac{\theta}{3} - i\sin^3\frac{\theta}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}\right)^3 = \cos\theta + i\sin\theta\) | B1 | si |
| \(\therefore \cos\theta = \cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}\sin^2\frac{\theta}{3}\) | M1 | FT (i) for sign error only |
| \(= \cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}\left(1 - \cos^2\frac{\theta}{3}\right)\) | A1 | |
| \(= 4\cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}\) | A1 | cao convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\cos\theta}{\cos\frac{\theta}{3}} = \frac{4\cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}}{\cos\frac{\theta}{3}} = 1\) | M1 | Substitution |
| \(4\cos^3\frac{\theta}{3} - 4\cos\frac{\theta}{3} = 0\) | A1 | Removing fraction |
| \(4\cos\frac{\theta}{3}\left(\cos^2\frac{\theta}{3} - 1\right) = 0\) | A1 | All three (including \(\pm 1\)) |
| \(\cos\frac{\theta}{3} = 0\) (not a possible solution in this equation) | ||
| or | A1 | |
| \(\cos\frac{\theta}{3} = \pm 1\) | ||
| When \(\cos\frac{\theta}{3} = 1\), \(\frac{\theta}{3} = 2n\pi\) | M1 | Use of general solution of \(\cos\theta\) |
| \(\therefore \theta = 6n\pi\) | ||
| When \(\cos\frac{\theta}{3} = -1\), \(\frac{\theta}{3} = \pi + 2n\pi\) | A1 | Either \(\theta\) |
| \(\therefore \theta = 3\pi + 6n\pi\) | A1 | |
| General solution: \(\theta = 3n\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\cos\theta}{\cos\frac{\theta}{3}} = 1\) | (B1) | |
| \(\cos\theta - \cos\frac{\theta}{3} = 0\) | (M1)(A1) | |
| Then, | ||
| \(-2\sin\frac{\theta + \frac{\theta}{3}}{2}\sin\frac{\theta - \frac{\theta}{3}}{2} = 0\) | (M1)(A1) | Both |
| Therefore, | ||
| \(\sin\frac{2\theta}{3} = 0\) or \(\sin\frac{\theta}{3} = 0\) | (A1) | Both |
| \(\frac{2\theta}{3} = n\pi\) or \(\frac{\theta}{3} = n\pi\) | (M1) | |
| \(\theta = \frac{3}{2}n\pi\) or \(\theta = 3n\pi\) | ||
| Odd multiple of \(\frac{3}{2}n\pi\) are not a solution because \(\cos\theta = 0\) | ||
| \(\theta = 3n\pi\) | (A1) |
**Part a) i)**
$\left(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}\right)^3$ | M1 | Unsimplified. Allow cis notation
$= \cos^3\frac{\theta}{3} + 3\cos^2\frac{\theta}{3}\left(i\sin\frac{\theta}{3}\right) + 3\cos\frac{\theta}{3}\left(i\sin\frac{\theta}{3}\right)^2 + \left(i\sin\frac{\theta}{3}\right)^3$ | A1 |
$= \cos^3\frac{\theta}{3} + 3i\cos^2\frac{\theta}{3}\sin\frac{\theta}{3} - 3\cos\frac{\theta}{3}\sin^2\frac{\theta}{3} - i\sin^3\frac{\theta}{3}$ | |
**Part a) ii)**
$\left(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}\right)^3 = \cos\theta + i\sin\theta$ | B1 | si
$\therefore \cos\theta = \cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}\sin^2\frac{\theta}{3}$ | M1 | FT (i) for sign error only
$= \cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}\left(1 - \cos^2\frac{\theta}{3}\right)$ | A1 |
$= 4\cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}$ | A1 | cao convincing
**Part b)**
METHOD 1:
$\frac{\cos\theta}{\cos\frac{\theta}{3}} = \frac{4\cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}}{\cos\frac{\theta}{3}} = 1$ | M1 | Substitution
$4\cos^3\frac{\theta}{3} - 4\cos\frac{\theta}{3} = 0$ | A1 | Removing fraction
$4\cos\frac{\theta}{3}\left(\cos^2\frac{\theta}{3} - 1\right) = 0$ | A1 | All three (including $\pm 1$)
$\cos\frac{\theta}{3} = 0$ (not a possible solution in this equation) | |
or | A1 |
$\cos\frac{\theta}{3} = \pm 1$ | |
When $\cos\frac{\theta}{3} = 1$, $\frac{\theta}{3} = 2n\pi$ | M1 | Use of general solution of $\cos\theta$
$\therefore \theta = 6n\pi$ | |
When $\cos\frac{\theta}{3} = -1$, $\frac{\theta}{3} = \pi + 2n\pi$ | A1 | Either $\theta$
$\therefore \theta = 3\pi + 6n\pi$ | A1 |
General solution: $\theta = 3n\pi$ | |
METHOD 2:
$\frac{\cos\theta}{\cos\frac{\theta}{3}} = 1$ | (B1) |
$\cos\theta - \cos\frac{\theta}{3} = 0$ | (M1)(A1) |
Then, | |
$-2\sin\frac{\theta + \frac{\theta}{3}}{2}\sin\frac{\theta - \frac{\theta}{3}}{2} = 0$ | (M1)(A1) | Both
Therefore, | |
$\sin\frac{2\theta}{3} = 0$ or $\sin\frac{\theta}{3} = 0$ | (A1) | Both
$\frac{2\theta}{3} = n\pi$ or $\frac{\theta}{3} = n\pi$ | (M1) |
$\theta = \frac{3}{2}n\pi$ or $\theta = 3n\pi$ | |
Odd multiple of $\frac{3}{2}n\pi$ are not a solution because $\cos\theta = 0$ | |
$\theta = 3n\pi$ | (A1) |
---\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Expand $\left(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3}\right)^3$.
\item Hence, by using de Moivre's theorem, show that $\cos\theta$ can be expressed as
$$4\cos^3\frac{\theta}{3} - 3\cos\frac{\theta}{3}.$$ [6]
\end{enumerate}
\item Hence, or otherwise, find the general solution of the equation $\frac{\cos\theta}{\cos\frac{\theta}{3}} = 1$. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q9 [12]}}