CAIE FP1 2012 November — Question 8 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeModulus-argument form conversions
DifficultyChallenging +1.2 This is a Further Maths FP1 question requiring modulus-argument form manipulation and binomial expansion with complex numbers. The first part uses standard half-angle identities (factor-formula technique), while the second part requires recognizing that expanding (1+z)^n and comparing imaginary parts yields the sum. It's a multi-step problem requiring insight to connect binomial theorem with De Moivre's theorem, making it moderately above average difficulty but still a recognizable FP1 technique.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae
8 Let \(z = \cos \theta + \mathrm { i } \sin \theta\). Show that $$1 + z = 2 \cos \frac { 1 } { 2 } \theta \left( \cos \frac { 1 } { 2 } \theta + \mathrm { i } \sin \frac { 1 } { 2 } \theta \right)$$ By considering \(( 1 + z ) ^ { n }\), where \(n\) is a positive integer, deduce the sum of the series $$\binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \ldots + \binom { n } { n } \sin n \theta$$
Question 8:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(1+z = 2\cos^2\frac{1}{2}\theta + 2i\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta\)M1A1 Re-writes
\(= 2\cos\frac{1}{2}\theta\left(\cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta\right)\) (AG)A1 Obtains result
\((1+z)^n = 1 + \binom{n}{1}z + \binom{n}{2}z^2 + \ldots + \binom{n}{n}z^n\)M1 Use of Binomial Theorem
\(\therefore \operatorname{Im}(1+z)^n = \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \ldots + \binom{n}{n}\sin n\theta\)M1, M1A1 Takes imaginary part and uses de Moivre
But \((1+z)^n = 2^n\cos^n\frac{1}{2}\theta\, e^{\frac{in}{2}\theta}\)B1 Applies initial result
\(\therefore \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \ldots + \binom{n}{n}\sin n\theta = 2^n\cos^n\frac{1}{2}\theta\sin\frac{n}{2}\theta\)A1 Equates imaginary parts 
## Question 8:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $1+z = 2\cos^2\frac{1}{2}\theta + 2i\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta$ | M1A1 | Re-writes |
| $= 2\cos\frac{1}{2}\theta\left(\cos\frac{1}{2}\theta + i\sin\frac{1}{2}\theta\right)$ (AG) | A1 | Obtains result |
| $(1+z)^n = 1 + \binom{n}{1}z + \binom{n}{2}z^2 + \ldots + \binom{n}{n}z^n$ | M1 | Use of Binomial Theorem |
| $\therefore \operatorname{Im}(1+z)^n = \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \ldots + \binom{n}{n}\sin n\theta$ | M1, M1A1 | Takes imaginary part and uses de Moivre |
| But $(1+z)^n = 2^n\cos^n\frac{1}{2}\theta\, e^{\frac{in}{2}\theta}$ | B1 | Applies initial result |
| $\therefore \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \ldots + \binom{n}{n}\sin n\theta = 2^n\cos^n\frac{1}{2}\theta\sin\frac{n}{2}\theta$ | A1 | Equates imaginary parts |

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8 Let $z = \cos \theta + \mathrm { i } \sin \theta$. Show that

$$1 + z = 2 \cos \frac { 1 } { 2 } \theta \left( \cos \frac { 1 } { 2 } \theta + \mathrm { i } \sin \frac { 1 } { 2 } \theta \right)$$

By considering $( 1 + z ) ^ { n }$, where $n$ is a positive integer, deduce the sum of the series

$$\binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \ldots + \binom { n } { n } \sin n \theta$$

\hfill \mbox{\textit{CAIE FP1 2012 Q8 [9]}}
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