| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP3 question using de Moivre's theorem to derive trigonometric identities. While it requires multiple steps and careful algebraic manipulation, the technique is well-practiced in FP3 courses. Part (i) is routine application of de Moivre, part (ii) is algebraic substitution, part (iii) involves setting cot 4θ = 0 to find roots (standard technique), and part (iv) requires recognizing complementary angles. The question is structured with clear guidance through each part, making it moderately challenging but within expected FP3 scope. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos4\theta+i\sin4\theta=c^4+4ic^3s-6c^2s^2-4ics^3+s^4\) | M1 | For using de Moivre with \(n=4\) |
| \(\Rightarrow\sin4\theta=4c^3s-4cs^3\) and \(\cos4\theta=c^4-6c^2s^2+s^4\) | A1 | For both expressions |
| \(\Rightarrow\tan4\theta=\dfrac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}\) | M1, A1 4 | For expressing \(\frac{\sin4\theta}{\cos4\theta}\) in terms of \(c\) and \(s\); for simplifying to correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cot4\theta=\dfrac{\cot^4\theta-6\cot^2\theta+1}{4\cot^3\theta-4\cot\theta}\) | B1 1 | For inverting (i) and using \(\cot\theta=\frac{1}{\tan\theta}\) or \(\tan\theta=\frac{1}{\cot\theta}\) AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cot4\theta=0\) | B1 | For putting \(\cot4\theta=0\) (can be awarded in (iv) if not earned here) |
| Put \(x=\cot^2\theta\) | B1 | For putting \(x=\cot^2\theta\) in the numerator of (ii) |
| \(\theta=\frac{1}{8}\pi\Rightarrow x^2-6x+1=0\) | B1 3 | For deducing quadratic from (ii) and \(\theta=\frac{1}{8}\pi\); OR for deducing \(\theta=\frac{1}{8}\pi\) from (ii) and quadratic |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4\theta=\frac{3}{2}\pi\) OR \(\frac{1}{2}(2n+1)\pi\) | M1 | For attempting to find another value of \(\theta\) |
| 2nd root is \(x=\cot^2\!\left(\frac{3}{8}\pi\right)\) | A1 | For the other root of the quadratic |
| \(\Rightarrow\cot^2\!\left(\frac{1}{8}\pi\right)+\cot^2\!\left(\frac{3}{8}\pi\right)=6\) | M1 | For using sum of roots of quadratic |
| \(\Rightarrow\csc^2\!\left(\frac{1}{8}\pi\right)+\csc^2\!\left(\frac{3}{8}\pi\right)=8\) | M1, A1 5 | For using \(\cot^2\theta+1=\csc^2\theta\); for correct value |
## Question 8:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos4\theta+i\sin4\theta=c^4+4ic^3s-6c^2s^2-4ics^3+s^4$ | M1 | For using de Moivre with $n=4$ |
| $\Rightarrow\sin4\theta=4c^3s-4cs^3$ and $\cos4\theta=c^4-6c^2s^2+s^4$ | A1 | For both expressions |
| $\Rightarrow\tan4\theta=\dfrac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}$ | M1, A1 **4** | For expressing $\frac{\sin4\theta}{\cos4\theta}$ in terms of $c$ and $s$; for simplifying to correct expression |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cot4\theta=\dfrac{\cot^4\theta-6\cot^2\theta+1}{4\cot^3\theta-4\cot\theta}$ | B1 **1** | For inverting (i) and using $\cot\theta=\frac{1}{\tan\theta}$ or $\tan\theta=\frac{1}{\cot\theta}$ AG |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cot4\theta=0$ | B1 | For putting $\cot4\theta=0$ (can be awarded in (iv) if not earned here) |
| Put $x=\cot^2\theta$ | B1 | For putting $x=\cot^2\theta$ in the numerator of (ii) |
| $\theta=\frac{1}{8}\pi\Rightarrow x^2-6x+1=0$ | B1 **3** | For deducing quadratic from (ii) and $\theta=\frac{1}{8}\pi$; OR for deducing $\theta=\frac{1}{8}\pi$ from (ii) and quadratic |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\theta=\frac{3}{2}\pi$ OR $\frac{1}{2}(2n+1)\pi$ | M1 | For attempting to find another value of $\theta$ |
| 2nd root is $x=\cot^2\!\left(\frac{3}{8}\pi\right)$ | A1 | For the other root of the quadratic |
| $\Rightarrow\cot^2\!\left(\frac{1}{8}\pi\right)+\cot^2\!\left(\frac{3}{8}\pi\right)=6$ | M1 | For using sum of roots of quadratic |
| $\Rightarrow\csc^2\!\left(\frac{1}{8}\pi\right)+\csc^2\!\left(\frac{3}{8}\pi\right)=8$ | M1, A1 **5** | For using $\cot^2\theta+1=\csc^2\theta$; for correct value |8 (i) Use de Moivre's theorem to find an expression for $\tan 4 \theta$ in terms of $\tan \theta$.\\
(ii) Deduce that $\cot 4 \theta = \frac { \cot ^ { 4 } \theta - 6 \cot ^ { 2 } \theta + 1 } { 4 \cot ^ { 3 } \theta - 4 \cot \theta }$.\\
(iii) Hence show that one of the roots of the equation $x ^ { 2 } - 6 x + 1 = 0$ is $\cot ^ { 2 } \left( \frac { 1 } { 8 } \pi \right)$.\\
(iv) Hence find the value of $\operatorname { cosec } ^ { 2 } \left( \frac { 1 } { 8 } \pi \right) + \operatorname { cosec } ^ { 2 } \left( \frac { 3 } { 8 } \pi \right)$, justifying your answer.
\hfill \mbox{\textit{OCR FP3 2007 Q8 [13]}}