OCR Further Pure Core 2 2022 June — Question 9 9 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive trigonometric identities
DifficultyChallenging +1.2 This is a standard Further Maths question using De Moivre's theorem to derive trigonometric identities. Part (a) requires binomial expansion and manipulation of exponentials (routine for FM students), while part (b) involves substituting a specific angle and algebraic manipulation to reach the given form. The question is structured with clear guidance ('show that') and follows a predictable pattern for this topic, making it moderately above average difficulty but well within the expected scope of FM Pure content.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials
9 In this question you must show detailed reasoning.
  1. Show that \(\operatorname { Re } \left( \mathrm { e } ^ { \mathrm { Ai } \theta } \left( \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 4 } \right) = a \cos 4 \theta \cos ^ { 4 } \theta\), where \(a\) is an integer to be determined.
  2. Hence show that \(\cos \frac { 1 } { 12 } \pi = \frac { 1 } { 2 } \sqrt [ 4 ] { \mathrm { b } + \mathrm { c } \sqrt { 3 } }\), where \(b\) and \(c\) are integers to be determined.
Question 9:
AnswerMarks Guidance
9(a) DR
e i e i 2 c o s o e    + − =
e 4 i c o s 4 i s i n 4    = +
( )
R e e 4 i ( e i e i 4 )    + −
R e ( ( c o s 4 i s i n 4 ) ( 2 c o s ) 4 )    = +
c o s 4 ( 2 c o s 4 )   = 
AnswerMarks
1 6 c o s 4 c o s 4   = (so a = 16)B1
M1
A1
AnswerMarks
[3]2.1
1.1
AnswerMarks
2.2aUse of DMT
Must equate terms to complete
demonstration
AnswerMarks
(b)DR
( i e e i ) 4 ( e i ) 4 4 ( i e 3 ) e i      + − = + −
6 ( e i ) 2 ( e i 2 ) 4 e i ( e i 3 ) ( e i ) 4      + − + − + −
e 4 i ( e i e i 4 )     + −
e 8 i 4 e 6 i 6 e 4 i 4 e 2 i 1     = + + + +
e 8 i c o s 8 i s i n 8    = + or R e ( e 8 i ) c o s 8   = etc
16cos4cos4=
cos8+4cos6+6cos4+4cos2+1
  
1 6 c o s c o 4 s  =  =
1 2 3 1 2
2    
c o s 4 c o s 6 c o s 4 c o s 1 + + + +
3 2 3 6
c o s  =
1 2
1 1 1 3  
4 4 0 6 4 1 − +  +  +  +
8 2 2 2
1  7  1 ( )
= 4 + 2 3 = 4 7 + 4 3
8 2 1 6
1 1
= 4 7 + 4 3 = 4 7 + 4 3
AnswerMarks
4 1 6 2*M1
dep*M1
dep*M1
dep*M1
dep*M1
A1
AnswerMarks
[6]3.1a
1.1
1.1
3.1a
1.1
AnswerMarks
2.2aExpands correct brackets using
binomial theorem. Terms can be
unsimplified but must have correct
numerical coefficients.
Use of Euler’s formula to convert
exponential form to trigonometric
form
Taking real parts
Choice of θ soi and substituted into
identity with their 16.
Gives correct numerical values to
n 
all c o s terms. Also dependent
6
on use of Euler’s formula and
choice of .
So b = 7 and c = 4 (can be
AnswerMarks
embedded). cao.eg ( e2i+1 )4 or ( z+z−1)4 or
 3 1 4
 +1+ i etc
 
 2 2 
if expansion seen in 9(a), must be
used in 9(b) to gain mark here
Question 9:
9 | (a) | DR
e i e i 2 c o s o e    + − =
e 4 i c o s 4 i s i n 4    = +
( )
R e e 4 i ( e i e i 4 )    + −
R e ( ( c o s 4 i s i n 4 ) ( 2 c o s ) 4 )    = +
c o s 4 ( 2 c o s 4 )   = 
1 6 c o s 4 c o s 4   = (so a = 16) | B1
M1
A1
[3] | 2.1
1.1
2.2a | Use of DMT
Must equate terms to complete
demonstration
(b) | DR
( i e e i ) 4 ( e i ) 4 4 ( i e 3 ) e i      + − = + −
6 ( e i ) 2 ( e i 2 ) 4 e i ( e i 3 ) ( e i ) 4      + − + − + −
e 4 i ( e i e i 4 )     + −
e 8 i 4 e 6 i 6 e 4 i 4 e 2 i 1     = + + + +
e 8 i c o s 8 i s i n 8    = + or R e ( e 8 i ) c o s 8   = etc
16cos4cos4=
cos8+4cos6+6cos4+4cos2+1
  
1 6 c o s c o 4 s  =  =
1 2 3 1 2
2    
c o s 4 c o s 6 c o s 4 c o s 1 + + + +
3 2 3 6

c o s  =
1 2
1 1 1 3  
4 4 0 6 4 1 − +  +  +  +
8 2 2 2
1  7  1 ( )
= 4 + 2 3 = 4 7 + 4 3
8 2 1 6
1 1
= 4 7 + 4 3 = 4 7 + 4 3
4 1 6 2 | *M1
dep*M1
dep*M1
dep*M1
dep*M1
A1
[6] | 3.1a
1.1
1.1
3.1a
1.1
2.2a | Expands correct brackets using
binomial theorem. Terms can be
unsimplified but must have correct
numerical coefficients.
Use of Euler’s formula to convert
exponential form to trigonometric
form
Taking real parts
Choice of θ soi and substituted into
identity with their 16.
Gives correct numerical values to
n 
all c o s terms. Also dependent
6
on use of Euler’s formula and
choice of .
So b = 7 and c = 4 (can be
embedded). cao. | eg ( e2i+1 )4 or ( z+z−1)4 or
 3 1 4
 +1+ i etc
 
 2 2 
if expansion seen in 9(a), must be
used in 9(b) to gain mark here
9 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Show that $\operatorname { Re } \left( \mathrm { e } ^ { \mathrm { Ai } \theta } \left( \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 4 } \right) = a \cos 4 \theta \cos ^ { 4 } \theta$, where $a$ is an integer to be determined.
\item Hence show that $\cos \frac { 1 } { 12 } \pi = \frac { 1 } { 2 } \sqrt [ 4 ] { \mathrm { b } + \mathrm { c } \sqrt { 3 } }$, where $b$ and $c$ are integers to be determined.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2022 Q9 [9]}}
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