CAIE FP1 2015 June — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSum geometric series with complex terms
DifficultyChallenging +1.8 This is a Further Maths question requiring De Moivre's theorem, binomial expansion, and manipulation of complex exponentials to derive a non-trivial trigonometric identity. While the techniques are standard for FP1, the multi-step proof and the need to connect real/imaginary parts with half-angle formulas elevates it above routine exercises. The second part follows naturally once the method is established.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)

6 Let \(z = \cos \theta + \mathrm { i } \sin \theta\). Use the binomial expansion of \(( 1 + z ) ^ { n }\), where \(n\) is a positive integer, to show that $$\binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta + \ldots + \binom { n } { n } \cos n \theta = 2 ^ { n } \cos ^ { n } \left( \frac { 1 } { 2 } \theta \right) \cos \left( \frac { 1 } { 2 } n \theta \right) - 1$$ Find $$\binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \ldots + \binom { n } { n } \sin n \theta$$

Question 6:
AnswerMarks Guidance
Working/AnswerMarks Notes
\((1+z)^n = 1 + \binom{n}{1}z + \binom{n}{2}z^2 + \cdots + \binom{n}{n}z^n\)B1
\(\text{Re}\{(1+z)^n\} = 1 + \binom{n}{1}\cos\theta + \binom{n}{2}\cos 2\theta + \cdots + \binom{n}{n}\cos n\theta\)M1A1
\(\text{Re}([1+\cos\theta] + i\sin\theta)^n = \text{Re}\!\left(2\cos^2\!\tfrac{1}{2}\theta + 2i\sin\tfrac{1}{2}\theta\cos\tfrac{1}{2}\theta\right)^n\)M1A1
\(= 2^n\cos^n\!\left(\tfrac{1}{2}\theta\right)\,\text{Re}\!\left(\cos\tfrac{1}{2}\theta + i\sin\tfrac{1}{2}\theta\right)^n\)A1
\(\Rightarrow \binom{n}{1}\cos\theta + \binom{n}{2}\cos 2\theta + \cdots + \binom{n}{n}\cos n\theta = 2^n\cos^n\!\tfrac{1}{2}\theta\cos\tfrac{1}{2}n\theta - 1\)A1 AG
\(\text{Im}\{(1+z)^n\} = \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \cdots + \binom{n}{n}\sin n\theta\)M1
\(\Rightarrow \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \cdots + \binom{n}{n}\sin n\theta = 2^n\cos^n\!\left(\tfrac{1}{2}\theta\right)\sin\!\left(\tfrac{1}{2}n\theta\right)\)A1
## Question 6:

| Working/Answer | Marks | Notes |
|---|---|---|
| $(1+z)^n = 1 + \binom{n}{1}z + \binom{n}{2}z^2 + \cdots + \binom{n}{n}z^n$ | B1 | |
| $\text{Re}\{(1+z)^n\} = 1 + \binom{n}{1}\cos\theta + \binom{n}{2}\cos 2\theta + \cdots + \binom{n}{n}\cos n\theta$ | M1A1 | |
| $\text{Re}([1+\cos\theta] + i\sin\theta)^n = \text{Re}\!\left(2\cos^2\!\tfrac{1}{2}\theta + 2i\sin\tfrac{1}{2}\theta\cos\tfrac{1}{2}\theta\right)^n$ | M1A1 | |
| $= 2^n\cos^n\!\left(\tfrac{1}{2}\theta\right)\,\text{Re}\!\left(\cos\tfrac{1}{2}\theta + i\sin\tfrac{1}{2}\theta\right)^n$ | A1 | |
| $\Rightarrow \binom{n}{1}\cos\theta + \binom{n}{2}\cos 2\theta + \cdots + \binom{n}{n}\cos n\theta = 2^n\cos^n\!\tfrac{1}{2}\theta\cos\tfrac{1}{2}n\theta - 1$ | A1 | AG |
| $\text{Im}\{(1+z)^n\} = \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \cdots + \binom{n}{n}\sin n\theta$ | M1 | |
| $\Rightarrow \binom{n}{1}\sin\theta + \binom{n}{2}\sin 2\theta + \cdots + \binom{n}{n}\sin n\theta = 2^n\cos^n\!\left(\tfrac{1}{2}\theta\right)\sin\!\left(\tfrac{1}{2}n\theta\right)$ | A1 | |

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6 Let $z = \cos \theta + \mathrm { i } \sin \theta$. Use the binomial expansion of $( 1 + z ) ^ { n }$, where $n$ is a positive integer, to show that

$$\binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta + \ldots + \binom { n } { n } \cos n \theta = 2 ^ { n } \cos ^ { n } \left( \frac { 1 } { 2 } \theta \right) \cos \left( \frac { 1 } { 2 } n \theta \right) - 1$$

Find

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

\hfill \mbox{\textit{CAIE FP1 2015 Q6 [9]}}