Question 9 - A-Level Maths

CAIE FP1 2013 November — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	November
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Proof by induction
Type	Prove de Moivre's theorem
Difficulty	Challenging +1.2 This is a two-part Further Maths question requiring a standard induction proof of de Moivre's theorem (which follows a well-established template using angle addition formulas) followed by applying binomial expansion and de Moivre to express sin^5θ in terms of multiple angles. While it requires multiple techniques and careful algebraic manipulation, both parts follow predictable methods taught explicitly in FP1 courses. The induction is more routine than creative, and the second part is a standard application rather than requiring novel insight.
Spec	4.01a Mathematical induction: construct proofs 4.02n Euler's formula: e^(itheta) = cos(theta) + isin(theta)4.02q De Moivre's theorem: multiple angle formulae

9 Prove by mathematical induction that, for every positive integer $n$, $$( \cos \theta + i \sin \theta ) ^ { n } = \cos n \theta + i \sin n \theta$$ Express $\sin ^ { 5 } \theta$ in the form $p \sin 5 \theta + q \sin 3 \theta + r \sin \theta$, where $p , q$ and $r$ are rational numbers to be determined.

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Question 9:

Part (induction):

Answer	Marks	Guidance
$H_k$: $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ for some $k$	B1
$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)(\cos k\theta + i\sin k\theta)$	M1
$= (\cos\theta\cos k\theta - \sin\theta\sin k\theta) + i(\sin\theta\cos k\theta + \cos\theta\sin k\theta)$	A1
$= \cos[k+1]\theta + i\sin[k+1]\theta \quad \therefore H_k \Rightarrow H_{k+1}$	A1
$H_1$ is trivially true, so true for all positive integers	A1	(5)

Part (de Moivre's theorem):

Answer	Marks
$z^n - \frac{1}{z^n} = (\cos n\theta + i\sin n\theta) - (\cos n\theta - i\sin n\theta) = 2i\sin n\theta$	B1

Part (binomial expansion):

Answer	Marks
$\left(z - \frac{1}{z}\right)^5 = (2i\sin\theta)^5 = 32i\sin^5\theta$	B1
$= z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}$	M1
$= \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)$	A1

Part (obtaining $\sin^5\theta$):

Answer	Marks	Guidance
$32i\sin^5\theta = 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta$	M1
$\sin^5\theta = \frac{1}{16}(\sin^5\theta - 5\sin 3\theta + 10\sin\theta)$	A1	(6)

# Question 9:

## Part (induction):
| $H_k$: $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ for some $k$ | B1 | |
|---|---|---|
| $(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)(\cos k\theta + i\sin k\theta)$ | M1 | |
| $= (\cos\theta\cos k\theta - \sin\theta\sin k\theta) + i(\sin\theta\cos k\theta + \cos\theta\sin k\theta)$ | A1 | |
| $= \cos[k+1]\theta + i\sin[k+1]\theta \quad \therefore H_k \Rightarrow H_{k+1}$ | A1 | |
| $H_1$ is trivially true, so true for all positive integers | A1 | **(5)** |

## Part (de Moivre's theorem):
| $z^n - \frac{1}{z^n} = (\cos n\theta + i\sin n\theta) - (\cos n\theta - i\sin n\theta) = 2i\sin n\theta$ | B1 | |
|---|---|---|

## Part (binomial expansion):
| $\left(z - \frac{1}{z}\right)^5 = (2i\sin\theta)^5 = 32i\sin^5\theta$ | B1 | |
|---|---|---|
| $= z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}$ | M1 | |
| $= \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)$ | A1 | |

## Part (obtaining $\sin^5\theta$):
| $32i\sin^5\theta = 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta$ | M1 | |
|---|---|---|
| $\sin^5\theta = \frac{1}{16}(\sin^5\theta - 5\sin 3\theta + 10\sin\theta)$ | A1 | **(6)** |

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9 Prove by mathematical induction that, for every positive integer $n$,

$$( \cos \theta + i \sin \theta ) ^ { n } = \cos n \theta + i \sin n \theta$$

Express $\sin ^ { 5 } \theta$ in the form $p \sin 5 \theta + q \sin 3 \theta + r \sin \theta$, where $p , q$ and $r$ are rational numbers to be determined.

\hfill \mbox{\textit{CAIE FP1 2013 Q9 [11]}}

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Q1 6 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 9 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR

Answer	Marks	Guidance
\(H_k\): \((\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta\) for some \(k\)	B1
\((\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)(\cos k\theta + i\sin k\theta)\)	M1
\(= (\cos\theta\cos k\theta - \sin\theta\sin k\theta) + i(\sin\theta\cos k\theta + \cos\theta\sin k\theta)\)	A1
\(= \cos[k+1]\theta + i\sin[k+1]\theta \quad \therefore H_k \Rightarrow H_{k+1}\)	A1
\(H_1\) is trivially true, so true for all positive integers	A1	(5)

Answer	Marks
\(\left(z - \frac{1}{z}\right)^5 = (2i\sin\theta)^5 = 32i\sin^5\theta\)	B1
\(= z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}\)	M1
\(= \left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)\)	A1

Answer	Marks	Guidance
\(32i\sin^5\theta = 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta\)	M1
\(\sin^5\theta = \frac{1}{16}(\sin^5\theta - 5\sin 3\theta + 10\sin\theta)\)	A1	(6)