OCR Further Pure Core 2 2019 June — Question 8 8 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyChallenging +1.2 This is a standard Further Maths question requiring systematic application of De Moivre's theorem and binomial expansion to express sin^6(θ) in terms of multiple angles, followed by substitution to find an exact value. While it involves multiple steps and algebraic manipulation, the technique is well-practiced in FP2 and follows a predictable pattern without requiring novel insight or particularly challenging problem-solving.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae
8 In this question you must show detailed reasoning.
  1. By writing \(\sin \theta\) in terms of \(\mathrm { e } ^ { \mathrm { i } \theta }\) and \(\mathrm { e } ^ { - \mathrm { i } \theta }\) show that $$\sin ^ { 6 } \theta = \frac { 1 } { 32 } ( 10 - 15 \cos 2 \theta + 6 \cos 4 \theta - \cos 6 \theta ) .$$
  2. Hence show that \(\sin \frac { 1 } { 8 } \pi = \frac { 1 } { 2 } \sqrt [ 6 ] { 20 - 14 \sqrt { 2 } }\).
Question 8:
AnswerMarks Guidance
8(a) DR
eiθ−e−iθ
sinθ=
2i
sin6θ= eiθ−e−iθ   6 =− 1 ( eiθ−e−iθ)6
 2i  64
e6i𝑖𝑖θ𝜃𝜃−6e4−iθ𝑖𝑖𝜃𝜃+615e2iθ−20+15e−2iθ−6e−4iθ+e−6iθ
(𝑒𝑒 −𝑒𝑒 ) =
e6iθ+e−6iθ−6 ( e4iθ+e−4iθ) +15 ( e2iθ+e−2iθ) −20
=2cos6θ−6×2cos4θ+15×2cos2θ−20
− 1 (2cos6θ−12cos4 6 θ+30cos2θ−20)
∴ 𝑐𝑐𝑠𝑠𝑛𝑛 𝑐𝑐 =
64
1
= (10−15cos2θ+6cos4θ−cos6θ)
AnswerMarks
32*B1
M1
M1
M1
dep*A1
AnswerMarks
[5]1.1a
2.1
1.1
2.1
AnswerMarks
1.1Raising expression for sin θ to
the power 6 and (2i)6 = -64.
Genuine attempt to use binomial
expansion with correct evaluated
binomial coefficients. Condone
sign errors
Collecting terms and using
eiφ+e−iφ=2cosφ at least once.
AnswerMarks
AG. Fully correct argumentCondone
i𝜃𝜃 −i𝜃𝜃
2i sin𝑐𝑐 = e −e
Allow use of for
𝑖𝑖𝑖𝑖 −𝑖𝑖𝑖𝑖
1st two M marks only 𝑒𝑒 +𝑒𝑒
𝑐𝑐𝑠𝑠𝑛𝑛𝑐𝑐 = 2𝑖𝑖
If i omitted from denominator
their expression for sin θ then
only this M mark can still be
awarded
π 1  2 − 2 
sin6 = 10−15× − (+6(0))
 
8 32 2 2 
π 1 ( )
sin = 6 20−15 2+ 2
8 64
1
= 6 20−14 2
AnswerMarks
2dep*M1
A1
AnswerMarks
[3]1.1
2.2aSubstitution and calculation of all
cos terms
AG Some intermediate working
AnswerMarks
must be seenTerms must be shown distinct
either in this line or in the form of
𝜋𝜋
cos𝑛𝑛8
Question 8:
8 | (a) | DR
eiθ−e−iθ
sinθ=
2i
sin6θ= eiθ−e−iθ   6 =− 1 ( eiθ−e−iθ)6
 2i  64
e6i𝑖𝑖θ𝜃𝜃−6e4−iθ𝑖𝑖𝜃𝜃+615e2iθ−20+15e−2iθ−6e−4iθ+e−6iθ
(𝑒𝑒 −𝑒𝑒 ) =
e6iθ+e−6iθ−6 ( e4iθ+e−4iθ) +15 ( e2iθ+e−2iθ) −20
=2cos6θ−6×2cos4θ+15×2cos2θ−20
− 1 (2cos6θ−12cos4 6 θ+30cos2θ−20)
∴ 𝑐𝑐𝑠𝑠𝑛𝑛 𝑐𝑐 =
64
1
= (10−15cos2θ+6cos4θ−cos6θ)
32 | *B1
M1
M1
M1
dep*A1
[5] | 1.1a
2.1
1.1
2.1
1.1 | Raising expression for sin θ to
the power 6 and (2i)6 = -64.
Genuine attempt to use binomial
expansion with correct evaluated
binomial coefficients. Condone
sign errors
Collecting terms and using
eiφ+e−iφ=2cosφ at least once.
AG. Fully correct argument | Condone
i𝜃𝜃 −i𝜃𝜃
2i sin𝑐𝑐 = e −e
Allow use of for
𝑖𝑖𝑖𝑖 −𝑖𝑖𝑖𝑖
1st two M marks only 𝑒𝑒 +𝑒𝑒
𝑐𝑐𝑠𝑠𝑛𝑛𝑐𝑐 = 2𝑖𝑖
If i omitted from denominator
their expression for sin θ then
only this M mark can still be
awarded
π 1  2 − 2 
sin6 = 10−15× − (+6(0))
 
8 32 2 2 
π 1 ( )
sin = 6 20−15 2+ 2
8 64
1
= 6 20−14 2
2 | dep*M1
A1
[3] | 1.1
2.2a | Substitution and calculation of all
cos terms
AG Some intermediate working
must be seen | Terms must be shown distinct
either in this line or in the form of
𝜋𝜋
cos𝑛𝑛8
8 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item By writing $\sin \theta$ in terms of $\mathrm { e } ^ { \mathrm { i } \theta }$ and $\mathrm { e } ^ { - \mathrm { i } \theta }$ show that

$$\sin ^ { 6 } \theta = \frac { 1 } { 32 } ( 10 - 15 \cos 2 \theta + 6 \cos 4 \theta - \cos 6 \theta ) .$$
\item Hence show that $\sin \frac { 1 } { 8 } \pi = \frac { 1 } { 2 } \sqrt [ 6 ] { 20 - 14 \sqrt { 2 } }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2019 Q8 [8]}}
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