| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| 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 substantial Further Maths question requiring multiple sophisticated techniques: deriving trig identities via de Moivre (standard FP2), proving a geometric series sum equals zero using complex exponentials (requires insight to recognize the geometric series structure), and Maclaurin series manipulation. Part (b) particularly requires recognizing the sum as a geometric series with complex common ratio—a non-trivial step. The multi-part nature and combination of techniques places this above average difficulty, though each individual part follows established FP2 methods. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos 5\theta + j\sin 5\theta = (\cos\theta + j\sin\theta)^5\) | M1 | Using de Moivre |
| \(= \cos^5\theta + 5\cos^4\theta\, j\sin\theta + 10\cos^3\theta\, j^2\sin^2\theta + 10\cos^2\theta\, j^3\sin^3\theta + 5\cos\theta\, j^4\sin^4\theta + j^5\sin^5\theta\) | M1 | Using binomial theorem appropriately |
| \(= \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta + j(\ldots)\) | A1 | Correct real part. Must evaluate powers of \(j\) |
| \(\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) | M1 | Equating real parts |
| \(= \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2\) | M1 | Replacing \(\sin^2\theta\) by \(1-\cos^2\theta\) |
| \(= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) | A1 | \(a=16,\ b=-20,\ c=5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C + jS = e^{j\theta} + e^{j(\theta+\frac{2\pi}{n})} + \ldots + e^{j(\theta+\frac{(2n-2)\pi}{n})}\) | M1 | Forming series \(C+jS\) as exponentials |
| Need not see whole series | A1 | |
| This is a G.P. | M1 | Attempting to sum finite or infinite G.P. |
| \(a = e^{j\theta},\ r = e^{j\frac{2\pi}{n}}\) | A1 | Correct \(a\), \(r\) used or stated, and \(n\) terms. Must see \(j\) |
| \(\text{Sum} = \dfrac{e^{j\theta}\!\left(1-\!\left(e^{j\frac{2\pi}{n}}\right)^n\right)}{1-e^{j\frac{2\pi}{n}}}\) | A1 | |
| Numerator \(= e^{j\theta}(1-e^{2\pi j})\) and \(e^{2\pi j}=1\) | ||
| so sum \(= 0\) | E1 | Convincing explanation that sum \(=0\) |
| \(\Rightarrow C=0\) and \(S=0\) | E1 | \(C=S=0\). Dep. on previous E1. Both E marks dep. on 5 marks above |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^t \approx 1 + t + \tfrac{1}{2}t^2\) | B1 | Ignore terms in higher powers |
| \(\dfrac{t}{e^t-1} \approx \dfrac{t}{t+\frac{1}{2}t^2}\) | M1 | Substituting Maclaurin series |
| \(\dfrac{t}{t+\frac{1}{2}t^2} = \dfrac{1}{1+\frac{1}{2}t} = (1+\tfrac{1}{2}t)^{-1} = 1-\tfrac{1}{2}t+\ldots\) | M1 | Suitable manipulation and use of binomial theorem |
| OR \(\dfrac{1}{1+\frac{1}{2}t} = \dfrac{1}{1+\frac{1}{2}t}\times\dfrac{1-\frac{1}{2}t}{1-\frac{1}{2}t} = \dfrac{1-\frac{1}{2}t}{1-\frac{1}{4}t^2}\) | M1 | |
| Hence \(\dfrac{t}{e^t-1} \approx 1-\tfrac{1}{2}t\) | A1 (ag) | |
| OR \((e^t-1)(1-\tfrac{1}{2}t) = (t+\tfrac{1}{2}t^2+\ldots)(1-\tfrac{1}{2}t)\) | M1 | Substituting Maclaurin series |
| \(\approx t + \text{terms in } t^3\) | M1, A1 | Correct expression; Multiplying out |
| \(\Rightarrow \dfrac{t}{e^t-1}\approx 1-\tfrac{1}{2}t\) | A1 | Convincing explanation |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 5\theta + j\sin 5\theta = (\cos\theta + j\sin\theta)^5$ | M1 | Using de Moivre |
| $= \cos^5\theta + 5\cos^4\theta\, j\sin\theta + 10\cos^3\theta\, j^2\sin^2\theta + 10\cos^2\theta\, j^3\sin^3\theta + 5\cos\theta\, j^4\sin^4\theta + j^5\sin^5\theta$ | M1 | Using binomial theorem appropriately |
| $= \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta + j(\ldots)$ | A1 | Correct real part. Must evaluate powers of $j$ |
| $\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$ | M1 | Equating real parts |
| $= \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2$ | M1 | Replacing $\sin^2\theta$ by $1-\cos^2\theta$ |
| $= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$ | A1 | $a=16,\ b=-20,\ c=5$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C + jS = e^{j\theta} + e^{j(\theta+\frac{2\pi}{n})} + \ldots + e^{j(\theta+\frac{(2n-2)\pi}{n})}$ | M1 | Forming series $C+jS$ as exponentials |
| Need not see whole series | A1 | |
| This is a G.P. | M1 | Attempting to sum finite or infinite G.P. |
| $a = e^{j\theta},\ r = e^{j\frac{2\pi}{n}}$ | A1 | Correct $a$, $r$ used or stated, and $n$ terms. Must see $j$ |
| $\text{Sum} = \dfrac{e^{j\theta}\!\left(1-\!\left(e^{j\frac{2\pi}{n}}\right)^n\right)}{1-e^{j\frac{2\pi}{n}}}$ | A1 | |
| Numerator $= e^{j\theta}(1-e^{2\pi j})$ and $e^{2\pi j}=1$ | | |
| so sum $= 0$ | E1 | Convincing explanation that sum $=0$ |
| $\Rightarrow C=0$ and $S=0$ | E1 | $C=S=0$. Dep. on previous E1. Both E marks dep. on 5 marks above |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^t \approx 1 + t + \tfrac{1}{2}t^2$ | B1 | Ignore terms in higher powers |
| $\dfrac{t}{e^t-1} \approx \dfrac{t}{t+\frac{1}{2}t^2}$ | M1 | Substituting Maclaurin series |
| $\dfrac{t}{t+\frac{1}{2}t^2} = \dfrac{1}{1+\frac{1}{2}t} = (1+\tfrac{1}{2}t)^{-1} = 1-\tfrac{1}{2}t+\ldots$ | M1 | Suitable manipulation and use of binomial theorem |
| OR $\dfrac{1}{1+\frac{1}{2}t} = \dfrac{1}{1+\frac{1}{2}t}\times\dfrac{1-\frac{1}{2}t}{1-\frac{1}{2}t} = \dfrac{1-\frac{1}{2}t}{1-\frac{1}{4}t^2}$ | M1 | |
| Hence $\dfrac{t}{e^t-1} \approx 1-\tfrac{1}{2}t$ | A1 (ag) | |
| OR $(e^t-1)(1-\tfrac{1}{2}t) = (t+\tfrac{1}{2}t^2+\ldots)(1-\tfrac{1}{2}t)$ | M1 | Substituting Maclaurin series |
| $\approx t + \text{terms in } t^3$ | M1, A1 | Correct expression; Multiplying out |
| $\Rightarrow \dfrac{t}{e^t-1}\approx 1-\tfrac{1}{2}t$ | A1 | Convincing explanation |
---2
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to find the constants $a , b , c$ in the identity
$$\cos 5 \theta \equiv a \cos ^ { 5 } \theta + b \cos ^ { 3 } \theta + c \cos \theta$$
\item Let
$$\begin{aligned}
C & = \cos \theta + \cos \left( \theta + \frac { 2 \pi } { n } \right) + \cos \left( \theta + \frac { 4 \pi } { n } \right) + \ldots + \cos \left( \theta + \frac { ( 2 n - 2 ) \pi } { n } \right) \\
\text { and } S & = \sin \theta + \sin \left( \theta + \frac { 2 \pi } { n } \right) + \sin \left( \theta + \frac { 4 \pi } { n } \right) + \ldots + \sin \left( \theta + \frac { ( 2 n - 2 ) \pi } { n } \right)
\end{aligned}$$
where $n$ is an integer greater than 1 .\\
By considering $C + \mathrm { j } S$, show that $C = 0$ and $S = 0$.
\item Write down the Maclaurin series for $\mathrm { e } ^ { t }$ as far as the term in $t ^ { 2 }$.
Hence show that, for $t$ close to zero,
$$\frac { t } { \mathrm { e } ^ { t } - 1 } \approx 1 - \frac { 1 } { 2 } t$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2010 Q2 [18]}}