| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on De Moivre's theorem with multiple parts requiring systematic application of known techniques. Parts (i) and (ii) are routine verifications, the sin^6θ derivation follows a well-established method using binomial expansion and the given identities, and finding cos^6θ uses the same approach. While it requires several steps and careful algebraic manipulation, it's a textbook-style question with no novel insights needed, making it moderately above average difficulty. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers |
7 Given that $z = \cos \theta + \mathrm { i } \sin \theta$, show that\\
(i) $z - \frac { 1 } { z } = 2 \mathrm { i } \sin \theta$.\\
(ii) $z ^ { n } + z ^ { - n } = 2 \cos n \theta$.
Hence show that
$$\sin ^ { 6 } \theta = \frac { 1 } { 32 } ( 10 - 15 \cos 2 \theta + 6 \cos 4 \theta - \cos 6 \theta )$$
Find a similar expression for $\cos ^ { 6 } \theta$, and hence express $\cos ^ { 6 } \theta - \sin ^ { 6 } \theta$ in the fom $a \cos 2 \theta + b \cos 6 \theta$.
\hfill \mbox{\textit{CAIE FP1 2002 Q7 [9]}}