| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Standard +0.8 This is a standard Further Maths question using de Moivre's theorem to derive a multiple angle formula, followed by a clever application to solve a quartic. Part (i) requires expanding (cos θ + i sin θ)^4, separating real/imaginary parts, and algebraic manipulation to get tan 4θ - routine for FM students. Part (ii) requires the insight to recognize the quartic matches the tan 4θ denominator/numerator structure, then solve tan 4θ = 1. While multi-step, this is a textbook FM technique with no exceptional difficulty beyond standard FM expectations. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \((c+is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4\) | B1 | SOI |
| Equate real and imaginary parts | M1 | |
| \(\cos 4\theta = c^4 - 6c^2s^2 + s^4\) | A1 | SOI |
| \(\sin 4\theta = 4c^3s - 4cs^3\) | A1 | SOI |
| \(\tan 4\theta = \frac{(4c^3s - 4cs^3) \div c^4}{(c^4 - 6c^2s^2 + s^4) \div c^4} \left(= \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}\right)\) | A1 | AG |
| Answer | Marks |
|---|---|
| \(\tan 4\theta = -1 \Rightarrow t^4 - 4t^3 - 6t^2 + 4t + 1 = 0\) | M1M1 |
| so \(4\theta = \frac{3\pi}{4}\left(\frac{7\pi}{4}, \frac{11\pi}{4}, \frac{15\pi}{4}\right)\) | M1 |
| \(t = \tan\frac{3\pi}{16}, \tan\frac{7\pi}{16}, \tan\frac{11\pi}{16}, \tan\frac{15\pi}{16}\). Allow \(\left(\frac{k}{4} - \frac{1}{16}\right)\pi\), \(k = 0,1,2,3\) oe | A1A1 |
## Question 7(i):
| $(c+is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4$ | B1 | SOI |
|---|---|---|
| Equate real and imaginary parts | M1 | |
| $\cos 4\theta = c^4 - 6c^2s^2 + s^4$ | A1 | SOI |
| $\sin 4\theta = 4c^3s - 4cs^3$ | A1 | SOI |
| $\tan 4\theta = \frac{(4c^3s - 4cs^3) \div c^4}{(c^4 - 6c^2s^2 + s^4) \div c^4} \left(= \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta}\right)$ | A1 | AG |
**Total: 5**
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## Question 7(ii):
| $\tan 4\theta = -1 \Rightarrow t^4 - 4t^3 - 6t^2 + 4t + 1 = 0$ | M1M1 | |
|---|---|---|
| so $4\theta = \frac{3\pi}{4}\left(\frac{7\pi}{4}, \frac{11\pi}{4}, \frac{15\pi}{4}\right)$ | M1 | |
| $t = \tan\frac{3\pi}{16}, \tan\frac{7\pi}{16}, \tan\frac{11\pi}{16}, \tan\frac{15\pi}{16}$. Allow $\left(\frac{k}{4} - \frac{1}{16}\right)\pi$, $k = 0,1,2,3$ oe | A1A1 | |
**Total: 5**
---7 (i) Use de Moivre's theorem to prove that
$$\tan 4 \theta = \frac { 4 \tan \theta - 4 \tan ^ { 3 } \theta } { 1 - 6 \tan ^ { 2 } \theta + \tan ^ { 4 } \theta } .$$
(ii) Hence find the solutions of the equation
$$t ^ { 4 } - 4 t ^ { 3 } - 6 t ^ { 2 } + 4 t + 1 = 0$$
giving your answers in the form $\tan k \pi$, where $k$ is a rational number.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q7 [10]}}