| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.8 This is a standard Further Maths question using De Moivre's theorem to derive trigonometric identities. Part (i) is routine verification, but part (ii) requires expanding (z - 1/z)^5 using the binomial theorem, then collecting terms and matching coefficients—a multi-step process requiring careful algebraic manipulation that goes beyond simple recall. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
**(i)** $z^n - \dfrac{1}{z^n} = (\cos n\theta + \mathrm{i}\sin n\theta) - (\cos[-n\theta] - \mathrm{i}\sin[-n\theta])$ **M1**
De Moivre's Thm. used for at least $z^n$
$= \cos n\theta + \mathrm{i}\sin n\theta - (\cos n\theta - \mathrm{i}\sin n\theta) = 2\mathrm{i}\sin n\theta$
Given answer obtained from 2 correct uses of de Moivre's Thm. and correct trig. **A1**
**[2]**
**(ii)** $\left(z - \dfrac{1}{z}\right)^5 = 32\mathrm{i}\sin^5\theta$ **M1**
$= z^5 - 5z^3 + 10z - \dfrac{10}{z} + \dfrac{5}{z^3} - \dfrac{1}{z^5}$ Use of binomial expansion **M1**
$= \left(z^5 - \dfrac{1}{z^5}\right) - 5\left(z^3 - \dfrac{1}{z^3}\right) + 10\left(z - \dfrac{1}{z}\right)$ Pairing up terms **M1**
$= 2\mathrm{i}\sin 5\theta - 10\mathrm{i}\sin 3\theta + 20\mathrm{i}\sin\theta$ Use of (i)'s result ($\times 3$) **M1**
$\Rightarrow \sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta$ **A1**
**[5]**5 Let $z = \cos \theta + \mathrm { i } \sin \theta$.\\
(i) Prove the result $z ^ { n } - \frac { 1 } { z ^ { n } } = 2 \mathrm { i } \sin n \theta$.\\
(ii) Use this result to express $\sin ^ { 5 } \theta$ in the form $A \sin 5 \theta + B \sin 3 \theta + C \sin \theta$, for constants $A , B$ and $C$ to be determined.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q5 [5]}}