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
$$\binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta + \ldots + \binom { n } { n } \sin n \theta$$
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Question 6:
Answer Marks
Guidance
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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## 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]}}