| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive trigonometric identities |
| Difficulty | Standard +0.8 This is a standard Further Maths FP3 question requiring de Moivre's theorem application and binomial expansion to derive a trigonometric identity, followed by a substitution to solve a polynomial equation. While it involves multiple steps and algebraic manipulation, it follows a well-established template that Further Maths students practice extensively. The techniques are routine for this level, though the algebra requires care. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((\cos 60) = \text{Re}(c+is)^6\) | M1 | For expanding (real part of) \((c+is)^6\) at least 4 terms and 1 evaluated binomial coefficient needed |
| \((\cos 60) = c^6 - 15c^4s^2 + 15c^2s^4 - s^6\) | A1 | For correct expansion |
| \(c^6 - 15c^4(1-c^2) + 15c^2(1-c^2)^2 - (1-c^2)^3\) | M1 | For using \(s^2 = 1 - c^2\) |
| \((\cos 60 =) 32c^6 - 48c^4 + 18c^2 - 1\) | A1 4 | For correct result AG |
| (ii) \(64x^3 - 96x^2 + 36x - 3 = 0 \Rightarrow \cos 60 = \frac{1}{2}\) | M1 | For obtaining a numerical value of \(\cos 60\) |
| \(\Rightarrow (\theta =) \frac{1}{18}\pi, \frac{5}{18}\pi, \frac{7}{18}\pi\) etc. | A1 | For any correct solution of \(\cos 60 = \frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| largest \(x\) requires smallest \(\theta\) | M1 | For stating or implying at least 2 values of \(\theta\) |
| \(\Rightarrow\) largest positive root is \(\cos \frac{1}{18}\pi\) | A1 4 | For identifying \(\cos \frac{1}{18}\pi\) AEF as the largest positive root from a list of 3 positive roots OR from general solution OR from consideration of the cosine function |
**(i)** $(\cos 60) = \text{Re}(c+is)^6$ | M1 | For expanding (real part of) $(c+is)^6$ at least 4 terms and 1 evaluated binomial coefficient needed
$(\cos 60) = c^6 - 15c^4s^2 + 15c^2s^4 - s^6$ | A1 | For correct expansion
$c^6 - 15c^4(1-c^2) + 15c^2(1-c^2)^2 - (1-c^2)^3$ | M1 | For using $s^2 = 1 - c^2$
$(\cos 60 =) 32c^6 - 48c^4 + 18c^2 - 1$ | A1 4 | For correct result AG
**(ii)** $64x^3 - 96x^2 + 36x - 3 = 0 \Rightarrow \cos 60 = \frac{1}{2}$ | M1 | For obtaining a numerical value of $\cos 60$
$\Rightarrow (\theta =) \frac{1}{18}\pi, \frac{5}{18}\pi, \frac{7}{18}\pi$ etc. | A1 | For any correct solution of $\cos 60 = \frac{1}{2}$
$\cos 60 = \frac{1}{2}$ has multiple roots
largest $x$ requires smallest $\theta$ | M1 | For stating or implying at least 2 values of $\theta$
$\Rightarrow$ largest positive root is $\cos \frac{1}{18}\pi$ | A1 4 | For identifying $\cos \frac{1}{18}\pi$ AEF as the largest positive root from a list of 3 positive roots OR from general solution OR from consideration of the cosine function5 (i) Use de Moivre's theorem to prove that
$$\cos 6 \theta = 32 \cos ^ { 6 } \theta - 48 \cos ^ { 4 } \theta + 18 \cos ^ { 2 } \theta - 1 .$$
(ii) Hence find the largest positive root of the equation
$$64 x ^ { 6 } - 96 x ^ { 4 } + 36 x ^ { 2 } - 3 = 0 ,$$
giving your answer in trigonometrical form.
\hfill \mbox{\textit{OCR FP3 2007 Q5 [8]}}