| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Standard +0.8 This is a standard Further Maths complex numbers question combining de Moivre's theorem, binomial expansion to express powers of sin/cos, and integration. Part (a) is routine proof, part (b) requires systematic algebraic manipulation using the result from (a), and part (c) is straightforward integration once (b) is complete. While multi-step and requiring several techniques, this follows a well-established template that Further Maths students practice extensively, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks |
|---|---|
| 8(a) | Obtains correct |
| Answer | Marks | Guidance |
|---|---|---|
| sinnθ | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| cosnθ+isinnθ | AO1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| and | AO2.1 | R1 |
| Answer | Marks |
|---|---|
| 8(b) | Selects the correctc os𝑛𝑛𝑛𝑛 |
| Answer | Marks | Guidance |
|---|---|---|
| z | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ignore LHS) | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| LHS) | AO1.1a | M1 |
| Obtains correct result | AO1.1b | A1 |
| Answer | Marks |
|---|---|
| 8(c) | Integrates their answer |
| Answer | Marks | Guidance |
|---|---|---|
| form ksinnθ | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| clearly | AO2.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS = 0/3 | AO2.1 | R1 |
| Total | 10 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Obtains correct
expression for in
terms of cosnθ a𝑛𝑛nd
𝑧𝑧
sinnθ | AO1.1b | B1 | zn =( cosθ+isinθ)n
=cosnθ+isinnθ
1
=( cosθ+isinθ)−n
zn
=cos (−nθ)+isin (−nθ)
=cosnθ−isinnθ
1
zn − =cosnθ+isinnθ−(cosnθ−isinnθ)
zn
=2isinnθ
Obtains correct
1
expression for in
zn
cosnθ
terms of and
sinnθ
Or expresses whole LHS
as
−2sin2nθ+2icosnθsinnθ
cosnθ+isinnθ | AO1.1b | B1
Completes a rigorous
argument (with all
intermediate steps) to
show the required
result, using properties
of sine and cosine
functions to obtain
results in terms of
and | AO2.1 | R1
--- 8(b) ---
8(b) | Selects the correctc os𝑛𝑛𝑛𝑛
sin𝑛𝑛𝑛𝑛
process by expanding
5
1
z−
z | AO3.1a | M1 | 15
z− =z5−5z3+10z−10z−1+5z−3−z−5
z
(2isinθ)5 =z5−z−5−5 ( z3−z−3) +10 ( z−z−1)
32isin5θ=2isin5θ−5(2isin3θ)+10(2isinθ)
1 5 5
sin5θ= sin5θ− sin3θ+ sinθ
16 16 8
Obtains three pairs of
1
terms in the form zn−
zn
(ignore LHS) | AO1.1a | M1
Replaces each pair with
the correct sinnθ
(ignore coefficients and
LHS) | AO1.1a | M1
Obtains correct result | AO1.1b | A1
--- 8(c) ---
8(c) | Integrates their answer
to part (b) correctly,
provided all terms in
integrand are of the
form ksinnθ | AO1.1a | M1 | π3 π3
1 5 5
∫ sin5θdθ= ∫ sin5θ− sin3θ+ sinθdθ
16 16 8
0 0
−1 5 5 π3
= cos5θ+ cos3θ− cosθ
80 48 8
0
1 5π 5 3π 5 π 1 5 5
=− cos + cos − cos −− cos0+ cos0− cos0
80 3 48 3 8 3 80 48 8
1 1 5 5 1 1 5 5
=− × + ×(−1)− × −− + −
80 2 48 8 2 80 48 8
53
=
480
Shows substitution
clearly | AO2.4 | M1
Completes a rigorous
argument to show the
required result.
NMS = 0/3 | AO2.1 | R1
Total | 10
Q | Marking Instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item If $z = \cos \theta + i \sin \theta$, use de Moivre's theorem to prove that
$$z^n - \frac{1}{z^n} = 2i \sin n\theta$$
[3 marks]
\item Express $\sin^5 \theta$ in terms of $\sin 5\theta$, $\sin 3\theta$ and $\sin \theta$
[4 marks]
\item Hence show that
$$\int_0^{\frac{\pi}{3}} \sin^5 \theta \, d\theta = \frac{53}{480}$$
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2019 Q8 [10]}}