OCR Further Pure Core 1 2020 November — Question 5 5 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2020
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive trigonometric identities
DifficultyStandard +0.8 This is a standard Further Maths question using De Moivre's theorem to derive a trigonometric identity. While it requires multiple steps (binomial expansion, substituting z = e^(iθ), using Euler's formula, and algebraic manipulation), the technique is well-practiced in FP1. It's harder than typical A-level pure questions due to the Further Maths content, but follows a predictable method without requiring novel insight.
Spec1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae
5 By expanding \(\left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 3 }\), where \(z = e ^ { \mathrm { i } \theta }\), show that \(4 \cos ^ { 3 } 2 \theta = \cos 6 \theta + 3 \cos 2 \theta\).
Question 5:
AnswerMarks
53
 1  3 1
z2 + = z6 +3z2 + +
 z2  z2 z6
3
1  1 
z2 + =2cos2θ⇒ z2 + =8cos32θ
z2  z2 
1 1
and z2 + =2cos2θand z6 + =2cos6θ
z2 z6
3
 1   1   1 
and z2 +  = z6 +  +3z2 + 
 z2   z6   z2 
⇒8cos32θ=2cos6θ+6cos2θ
⇒4cos32θ=cos6θ+3cos2θ
AnswerMarks
AGM1
A1
M1
A1
AnswerMarks
A11.1
1.1
3.1a
2.1
AnswerMarks
1.1Use of binomial
use of De Moivre for any one term
[5]
Question 5:
5 | 3
 1  3 1
z2 + = z6 +3z2 + +

 z2  z2 z6
3
1  1 
z2 + =2cos2θ⇒ z2 + =8cos32θ

z2  z2 
1 1
and z2 + =2cos2θand z6 + =2cos6θ
z2 z6
3
 1   1   1 
and z2 +  = z6 +  +3z2 + 
 z2   z6   z2 
⇒8cos32θ=2cos6θ+6cos2θ
⇒4cos32θ=cos6θ+3cos2θ
AG | M1
A1
M1
A1
A1 | 1.1
1.1
3.1a
2.1
1.1 | Use of binomial
use of De Moivre for any one term
[5]
5 By expanding $\left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 3 }$, where $z = e ^ { \mathrm { i } \theta }$, show that $4 \cos ^ { 3 } 2 \theta = \cos 6 \theta + 3 \cos 2 \theta$.

\hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q5 [5]}}
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Q1 2 Q2 3 Q3 5 Q4 4 Q5 5 Q6 5 Q7 5 Q8 10 Q9 9 Q10 13 Q11 8 Q12 6