| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Express roots in trigonometric form |
| Difficulty | Challenging +1.8 Part (a) requires systematic application of de Moivre's theorem with binomial expansion and algebraic manipulation to derive a tan(5θ) identity—a standard Further Maths technique but requiring careful execution. Part (b) involves recognizing that tan²(π/5) and tan²(2π/5) are roots of a quadratic, connecting the tan(5θ) result to finding these specific values—this requires non-trivial insight linking the identity to roots of unity and forming the quadratic. The multi-step reasoning and connection between parts elevates this above routine Further Maths exercises. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta\) | B1 | |
| \(= c^5 + 5ic^4s - 10ic^3s^2 - 10ic^2s^3 + 5cs^4 + is^5\) | M1A1 | Binomial expansion |
| \(\tan 5\theta = \frac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}\) | M1 | |
| \(\tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}\) | A1 | AG Division of each term by \(c^5\) clearly stated |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Roots of \(\tan 5\theta = 0\) are \(\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}\) | B1 | Allow inclusion of 0 here |
| \(t^4 - 10t^2 + 5 = 0\) has roots \(\tan\frac{\pi}{5}, \tan\frac{2\pi}{5}, \tan\frac{3\pi}{5}, \tan\frac{4\pi}{5}\) | B1 | |
| \(\left(t - \tan\frac{2\pi}{5}\right)\left(t - \tan\frac{3\pi}{5}\right)\left(t - \tan\frac{4\pi}{5}\right) = 0\) | M1 | |
| Since \(\tan\frac{4\pi}{5} = -\tan\frac{\pi}{5}\) and \(\tan\frac{3\pi}{5} = -\tan\frac{2\pi}{5}\), | M1 | |
| \(\left(t^2 - \tan^2\frac{\pi}{5}\right)\left(t^2 - \tan^2\frac{2\pi}{5}\right) = 0\) | ||
| So roots of \(x^2 - 10x + 5 = 0\) are \(\tan^2\left(\frac{\pi}{5}\right)\) and \(\tan^2\left(\frac{2\pi}{5}\right)\) | A1 | AG |
| Total: 5 |
# Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$ | B1 | |
| $= c^5 + 5ic^4s - 10ic^3s^2 - 10ic^2s^3 + 5cs^4 + is^5$ | M1A1 | Binomial expansion |
| $\tan 5\theta = \frac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}$ | M1 | |
| $\tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}$ | A1 | AG Division of each term by $c^5$ clearly stated |
| **Total: 5** | | |
# Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Roots of $\tan 5\theta = 0$ are $\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}$ | B1 | Allow inclusion of 0 here |
| $t^4 - 10t^2 + 5 = 0$ has roots $\tan\frac{\pi}{5}, \tan\frac{2\pi}{5}, \tan\frac{3\pi}{5}, \tan\frac{4\pi}{5}$ | B1 | |
| $\left(t - \tan\frac{2\pi}{5}\right)\left(t - \tan\frac{3\pi}{5}\right)\left(t - \tan\frac{4\pi}{5}\right) = 0$ | M1 | |
| Since $\tan\frac{4\pi}{5} = -\tan\frac{\pi}{5}$ and $\tan\frac{3\pi}{5} = -\tan\frac{2\pi}{5}$, | M1 | |
| $\left(t^2 - \tan^2\frac{\pi}{5}\right)\left(t^2 - \tan^2\frac{2\pi}{5}\right) = 0$ | | |
| So roots of $x^2 - 10x + 5 = 0$ are $\tan^2\left(\frac{\pi}{5}\right)$ and $\tan^2\left(\frac{2\pi}{5}\right)$ | A1 | AG |
| **Total: 5** | | |6 (a) Using de Moivre's theorem, show that
$$\tan 5 \theta = \frac { 5 \tan \theta - 10 \tan ^ { 3 } \theta + \tan ^ { 5 } \theta } { 1 - 10 \tan ^ { 2 } \theta + 5 \tan ^ { 4 } \theta } .$$
(b) Hence show that the equation $x ^ { 2 } - 10 x + 5 = 0$ has roots $\tan ^ { 2 } \left( \frac { 1 } { 5 } \pi \right)$ and $\tan ^ { 2 } \left( \frac { 2 } { 5 } \pi \right)$.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [10]}}