CAIE Further Paper 2 2021 June — Question 4 7 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive tan/cot identities
DifficultyChallenging +1.2 This is a standard Further Maths technique combining binomial expansion with de Moivre's theorem to derive a multiple angle identity. While it requires careful algebraic manipulation and understanding of the method (substituting z = cos θ + i sin θ, expanding both expressions, using z + 1/z = 2cos θ and z - 1/z = 2i sin θ), it follows a well-established procedure taught explicitly in Further Pure courses. The question guides students by specifying which expressions to expand and what to show, making it more routine than exploratory.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n4.02q De Moivre's theorem: multiple angle formulae
4 By considering the binomial expansions of \(\left( z + \frac { 1 } { z } \right) ^ { 5 }\) and \(\left( z - \frac { 1 } { z } \right) ^ { 5 }\), where \(z = \cos \theta + \mathrm { i } \sin \theta\), use de Moivre's theorem to show that $$\tan ^ { 5 } \theta = \frac { \sin 5 \theta - \mathrm { a } \sin 3 \theta + \mathrm { b } \sin \theta } { \cos 5 \theta + \mathrm { a } \cos 3 \theta + \mathrm { b } \cos \theta }$$ where \(a\) and \(b\) are integers to be determined.
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(z - z^{-1} = 2i\sin\theta\) and \(z + z^{-1} = 2\cos\theta\)B1 Use of \(z - z^{-1} = 2i\sin\theta\) and \(z + z^{-1} = 2\cos\theta\)
\((z-z^{-1})^5 = (z^5-z^{-5}) - 5(z^3-z^{-3}) + 10(z-z^{-1})\)M1 A1 Expands and groups. A0 if grouping not shown
\((z+z^{-1})^5 = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})\)M1 A1 Expands and groups. A0 if grouping not shown
\(\frac{2^5 i\sin^5\theta}{2^5\cos^5\theta} = \frac{2i\sin 5\theta - 5(2i\sin 3\theta)+10(2i\sin\theta)}{2\cos 5\theta + 5(2\cos 3\theta)+10(2\cos\theta)}\)M1 Substitutes \(z^n + z^{-n} = 2\cos n\theta\) and \(z^n - z^{-n} = 2i\sin n\theta\)
\(\tan^5\theta = \frac{\sin 5\theta - 5\sin 3\theta + 10\sin\theta}{\cos 5\theta + 5\cos 3\theta + 10\cos\theta}\)A1 Justifies cancellation of constants
Total: 7 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$ | B1 | Use of $z - z^{-1} = 2i\sin\theta$ and $z + z^{-1} = 2\cos\theta$ |
| $(z-z^{-1})^5 = (z^5-z^{-5}) - 5(z^3-z^{-3}) + 10(z-z^{-1})$ | M1 A1 | Expands and groups. A0 if grouping not shown |
| $(z+z^{-1})^5 = (z^5+z^{-5}) + 5(z^3+z^{-3}) + 10(z+z^{-1})$ | M1 A1 | Expands and groups. A0 if grouping not shown |
| $\frac{2^5 i\sin^5\theta}{2^5\cos^5\theta} = \frac{2i\sin 5\theta - 5(2i\sin 3\theta)+10(2i\sin\theta)}{2\cos 5\theta + 5(2\cos 3\theta)+10(2\cos\theta)}$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$ and $z^n - z^{-n} = 2i\sin n\theta$ |
| $\tan^5\theta = \frac{\sin 5\theta - 5\sin 3\theta + 10\sin\theta}{\cos 5\theta + 5\cos 3\theta + 10\cos\theta}$ | A1 | Justifies cancellation of constants |
| **Total: 7** | | |

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4 By considering the binomial expansions of $\left( z + \frac { 1 } { z } \right) ^ { 5 }$ and $\left( z - \frac { 1 } { z } \right) ^ { 5 }$, where $z = \cos \theta + \mathrm { i } \sin \theta$, use de Moivre's theorem to show that

$$\tan ^ { 5 } \theta = \frac { \sin 5 \theta - \mathrm { a } \sin 3 \theta + \mathrm { b } \sin \theta } { \cos 5 \theta + \mathrm { a } \cos 3 \theta + \mathrm { b } \cos \theta }$$

where $a$ and $b$ are integers to be determined.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q4 [7]}}
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Q1 7 Q2 7 Q3 10 Q4 7 Q5 10 Q6 10 Q7 11 Q8 13