| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| 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 FP3 question requiring de Moivre's theorem application and binomial expansion to derive a trigonometric identity, followed by a straightforward inequality proof using the derived result. While it involves multiple steps and algebraic manipulation, the techniques are well-practiced in Further Maths courses with no novel insight required. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6 = \cos^6\theta + 6i\sin\theta\cos^5\theta - 15\sin^2\theta\cos^4\theta - 20i\sin^3\theta\cos^3\theta + 15\sin^4\theta\cos^2\theta + 6i\sin^5\theta\cos\theta - \sin^6\theta\) | B1 | Use de Moivre; or \(\sin 6\theta = \text{Im}(\cos\theta+i\sin\theta)^6\) |
| All terms correct | B1 | |
| \(\sin 6\theta = 6\sin\theta\cos^5\theta - 20\sin^3\theta\cos^3\theta + 6\sin^5\theta\cos\theta\) | M1 | Compare imaginary parts |
| \(= \cos\theta(6\sin\theta(1-\sin^2\theta)^2 - 20\sin^3\theta(1-\sin^2\theta)+6\sin^5\theta)\) | M1 | Take out factor of \(\cos\theta\) and give other factor in terms of \(\sin\theta\) only |
| \(= \cos\theta(32\sin^5\theta - 32\sin^3\theta + 6\sin\theta)\) | A1 | ag Convincingly shown, having been explicit about taking imaginary parts; must have \(\sin 6\theta = \cdots\) 'final line' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\sin 6\theta}{\sin 2\theta} = \dfrac{\cos\theta(32\sin^5\theta - 32\sin^3\theta + 6\sin\theta)}{2\sin\theta\cos\theta}\) | M1 | |
| \(= 16\sin^4\theta - 16\sin^2\theta + 3\) | M1 | Complete the square |
| \(= 4(2\sin^2\theta - 1)^2 - 1\) | A1 | |
| \(\therefore \dfrac{\sin 6\theta}{\sin 2\theta} \geq -1\) | M1 | Deduces lower bound |
| \(0 \leq 2\sin^2\theta \leq 2 \therefore (2\sin^2\theta-1)^2 \leq 1\) | M1 | Deduces upper bound |
| \(\therefore 4(2\sin^2\theta-1)^2 - 1 \leq 3 \Rightarrow -1 \leq \dfrac{\sin 6\theta}{\sin 2\theta} \leq 3\) | SC if none of marks 2–5 gained then SC M1A1 for any valid method of deducing upper and lower bounds | |
| But upper bound attained \(\Rightarrow \sin^2\theta = 0\) or \(1 \Rightarrow \sin 2\theta = 0\) | M1 | Dep on showing valid method for UB\(\leq 3\); or independent proof that not equal to 3 |
| So \(\sin 2\theta \neq 0 \Rightarrow -1 \leq \dfrac{\sin 6\theta}{\sin 2\theta} < 3\) | A1 | Full convincing overall argument |
# Question 7:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6 = \cos^6\theta + 6i\sin\theta\cos^5\theta - 15\sin^2\theta\cos^4\theta - 20i\sin^3\theta\cos^3\theta + 15\sin^4\theta\cos^2\theta + 6i\sin^5\theta\cos\theta - \sin^6\theta$ | B1 | Use de Moivre; or $\sin 6\theta = \text{Im}(\cos\theta+i\sin\theta)^6$ |
| All terms correct | B1 | |
| $\sin 6\theta = 6\sin\theta\cos^5\theta - 20\sin^3\theta\cos^3\theta + 6\sin^5\theta\cos\theta$ | M1 | Compare imaginary parts |
| $= \cos\theta(6\sin\theta(1-\sin^2\theta)^2 - 20\sin^3\theta(1-\sin^2\theta)+6\sin^5\theta)$ | M1 | Take out factor of $\cos\theta$ and give other factor in terms of $\sin\theta$ only |
| $= \cos\theta(32\sin^5\theta - 32\sin^3\theta + 6\sin\theta)$ | A1 | ag Convincingly shown, having been explicit about taking imaginary parts; must have $\sin 6\theta = \cdots$ 'final line' |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\sin 6\theta}{\sin 2\theta} = \dfrac{\cos\theta(32\sin^5\theta - 32\sin^3\theta + 6\sin\theta)}{2\sin\theta\cos\theta}$ | M1 | |
| $= 16\sin^4\theta - 16\sin^2\theta + 3$ | M1 | Complete the square |
| $= 4(2\sin^2\theta - 1)^2 - 1$ | A1 | |
| $\therefore \dfrac{\sin 6\theta}{\sin 2\theta} \geq -1$ | M1 | Deduces lower bound |
| $0 \leq 2\sin^2\theta \leq 2 \therefore (2\sin^2\theta-1)^2 \leq 1$ | M1 | Deduces upper bound |
| $\therefore 4(2\sin^2\theta-1)^2 - 1 \leq 3 \Rightarrow -1 \leq \dfrac{\sin 6\theta}{\sin 2\theta} \leq 3$ | | SC if none of marks 2–5 gained then SC M1A1 for any valid method of deducing upper and lower bounds |
| But upper bound attained $\Rightarrow \sin^2\theta = 0$ or $1 \Rightarrow \sin 2\theta = 0$ | M1 | Dep on showing valid method for UB$\leq 3$; or independent proof that not equal to 3 |
| So $\sin 2\theta \neq 0 \Rightarrow -1 \leq \dfrac{\sin 6\theta}{\sin 2\theta} < 3$ | A1 | Full convincing overall argument |(i) Use de Moivre's theorem to show that
$$\sin 6 \theta \equiv \cos \theta \left( 6 \sin \theta - 32 \sin ^ { 3 } \theta + 32 \sin ^ { 5 } \theta \right)$$
(ii) Hence show that, for $\sin 2 \theta \neq 0$,
$$- 1 \leqslant \frac { \sin 6 \theta } { \sin 2 \theta } < 3$$
\hfill \mbox{\textit{OCR FP3 2016 Q7 [12]}}