CAIE FP1 2011 November — Question 5 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyStandard +0.3 This is a standard Further Maths question following a well-established procedure: apply de Moivre's theorem to express cos^4θ as a sum of multiple angle terms, then integrate. The method is algorithmic (binomial expansion of (e^{iθ} + e^{-iθ})^4, collect terms) and the integration is straightforward once the expression is obtained. While it requires knowledge of complex numbers and de Moivre's theorem (making it harder than typical A-level), it's a routine textbook exercise for FP1 students with no novel problem-solving required.
Spec1.08b Integrate x^n: where n != -1 and sums4.02q De Moivre's theorem: multiple angle formulae

5 Use de Moivre's theorem to express \(\cos ^ { 4 } \theta\) in the form $$a \cos 4 \theta + b \cos 2 \theta + c$$ where \(a , b , c\) are constants to be found. Hence evaluate $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 4 } \theta d \theta$$ leaving your answer in terms of \(\pi\).

Question 5:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\((z+z^{-1})^4 = (z^4+z^{-4})+4(z^2+z^{-2})+6\), where \(z=\cos\theta+i\sin\theta\)M1A1 Binomial expansion and groups
\((2\cos\theta)^4 = 2\cos 4\theta + 8\cos 2\theta + 6\)M1 Uses de Moivre's theorem
\(\cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}\)A1 Simplifies; Part marks: 4
\(\int_0^{\pi/4}\cos^4\theta\,d\theta = \int_0^{\pi/4}\left(\frac{1}{8}\cos 4\theta+\frac{1}{2}\cos 2\theta+\frac{3}{8}\right)d\theta\)M1 Integrates result
\(= \left[\frac{\sin 4\theta}{32}+\frac{\sin 2\theta}{4}+\frac{3\theta}{8}\right]_0^{\pi/4}\)A1 Correctly
\(= \frac{1}{4}+\frac{3\pi}{32}\)A1 Evaluates; Part marks: 3; Total: [7]
## Question 5:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $(z+z^{-1})^4 = (z^4+z^{-4})+4(z^2+z^{-2})+6$, where $z=\cos\theta+i\sin\theta$ | M1A1 | Binomial expansion and groups |
| $(2\cos\theta)^4 = 2\cos 4\theta + 8\cos 2\theta + 6$ | M1 | Uses de Moivre's theorem |
| $\cos^4\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$ | A1 | Simplifies; Part marks: 4 |
| $\int_0^{\pi/4}\cos^4\theta\,d\theta = \int_0^{\pi/4}\left(\frac{1}{8}\cos 4\theta+\frac{1}{2}\cos 2\theta+\frac{3}{8}\right)d\theta$ | M1 | Integrates result |
| $= \left[\frac{\sin 4\theta}{32}+\frac{\sin 2\theta}{4}+\frac{3\theta}{8}\right]_0^{\pi/4}$ | A1 | Correctly |
| $= \frac{1}{4}+\frac{3\pi}{32}$ | A1 | Evaluates; Part marks: 3; **Total: [7]** |

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5 Use de Moivre's theorem to express $\cos ^ { 4 } \theta$ in the form

$$a \cos 4 \theta + b \cos 2 \theta + c$$

where $a , b , c$ are constants to be found.

Hence evaluate

$$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { 4 } \theta d \theta$$

leaving your answer in terms of $\pi$.

\hfill \mbox{\textit{CAIE FP1 2011 Q5 [7]}}