| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths application of de Moivre's theorem requiring binomial expansion of (cos θ + i sin θ)^8, collecting real parts, and expressing in terms of cos θ only. Part (i) uses the Pythagorean identity substitution, and part (ii) involves recognizing that cos(π/8) satisfies cos(8θ)=0 and substituting into the derived polynomial. While multi-step and requiring careful algebra, this follows a well-established template taught in FP1 with no novel insight required. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks |
|---|---|
| Write \(c = \cos\theta, \quad s = \sin\theta, \quad c_n = \cos(n\theta)\) | |
| \(c_8 + is_8 = (c + is)^8 \Rightarrow c_8 = c^8 - 28c^6s^2 + 70c^4s^4 - 28c^2s^6 + s^8\) | M1A1 |
| \(\Rightarrow c_8 = c^8 - 28c^6(1-c^2) + 70c^4(1-2c^2+c^4) - 28c^2(1-3c^2+3c^4-c^6) + (1-4c^2+6c^4-4c^6+c^8)\) | M1A1 |
| \(\Rightarrow c_8 = 128c^8 - 256c^6 + 160c^4 - 32c^2 + 1\) (*) | A1 |
| (i) \(\theta \to \pi/2 - \theta\) in (*) leads to: | |
| \(c_8 = 128s^8 - 256s^6 + 160s^4 - 32s^2 + 1\) | M1A1 |
| (ii) From (*), \(x = \cos^2\pi/8 \Rightarrow 32(4x^4 - 8x^3 + 5x^2 - x) + 1 = \cos\pi = -1\) | M1M1 |
| \(\Rightarrow 4x^4 - 8x^3 + 5x^2 - x = -1/16\) | A1 |
Write $c = \cos\theta, \quad s = \sin\theta, \quad c_n = \cos(n\theta)$ | |
$c_8 + is_8 = (c + is)^8 \Rightarrow c_8 = c^8 - 28c^6s^2 + 70c^4s^4 - 28c^2s^6 + s^8$ | M1A1 |
$\Rightarrow c_8 = c^8 - 28c^6(1-c^2) + 70c^4(1-2c^2+c^4) - 28c^2(1-3c^2+3c^4-c^6) + (1-4c^2+6c^4-4c^6+c^8)$ | M1A1 |
$\Rightarrow c_8 = 128c^8 - 256c^6 + 160c^4 - 32c^2 + 1$ (*) | A1 |
**(i)** $\theta \to \pi/2 - \theta$ in (*) leads to: | |
$c_8 = 128s^8 - 256s^6 + 160s^4 - 32s^2 + 1$ | M1A1 |
**(ii)** From (*), $x = \cos^2\pi/8 \Rightarrow 32(4x^4 - 8x^3 + 5x^2 - x) + 1 = \cos\pi = -1$ | M1M1 |
$\Rightarrow 4x^4 - 8x^3 + 5x^2 - x = -1/16$ | A1 |10 Use de Moivre's theorem to express $\cos 8 \theta$ as a polynomial in $\cos \theta$.
Hence\\
(i) express $\cos 8 \theta$ as a polynomial in $\sin \theta$,\\
(ii) find the exact value of
$$4 x ^ { 4 } - 8 x ^ { 3 } + 5 x ^ { 2 } - x$$
where $x = \cos ^ { 2 } \left( \frac { 1 } { 8 } \pi \right)$.
\hfill \mbox{\textit{CAIE FP1 2008 Q10 [10]}}