AQA Paper 3 2018 June — Question 8 9 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyStandard +0.3 Part (a) is a routine trigonometric identity proof using standard identities (sin 2x = 2sin x cos x and 1 + tan²x = sec²x), requiring 2-3 steps. Part (b) applies the proven identity with substitution (recognizing the pattern with 2θ instead of x) and integrating -cos⁴(2θ), which is straightforward once the setup is recognized. This is slightly easier than average as it's a standard 'prove then apply' structure with well-practiced techniques.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)
  1. Prove the identity \(\frac{\sin 2x}{1 + \tan^2 x} = 2\sin x \cos^3 x\) [3 marks]
  2. Hence find \(\int \frac{4\sin 4\theta}{1 + \tan^2 2\theta} d\theta\) [6 marks]
Question 8:
AnswerMarks Guidance
8(a)Recalls a correct trig identity, which
could lead to a correct answerAO1.2 B1
sin2x
1+tan2 x
2sinxcosx
≡
1+tan2 x
2sinxcosx
≡
sec2 x
≡2sinxcosxcos2 x
≡2sinxcos3 x
(≡RHS )
Demonstrates a strategy for
proving the identity, eg by
converting all the terms on the LHS
AnswerMarks Guidance
to cos and sin.AO3.1a M1
Concludes a rigorous mathematical
argument to prove given identity
AnswerMarks Guidance
AGAO2.1 R1
AnswerMarks Guidance
8(b)Uses identity to write integrand in
the form asin2θcos32θAO1.1a M1
∫ dθ=∫8sin2θcos32θdθ
1+tan22θ
Let u=cos2θ
du 1 du
then =−2sin2θ⇒sin2θ=−
dθ 2dθ
du
I =−4∫u3 dθ
dθ
=−4∫u3du
=−u4+c
=−cos42θ+c
Correctly writes integrand as
AnswerMarks Guidance
8sin2θcos32θAO1.1b A1
Selects an appropriate method for
integrating,
e.g. substitution u =cos2θ,
or by inspection PI by sight of
AnswerMarks Guidance
cos42θAO3.1a M1
k∫u3du
Obtains correctly
PI by solution in form kcos42θ, if
AnswerMarks Guidance
by inspectionAO1.1a M1
Obtains−u4 or −cos42θOE
AnswerMarks Guidance
Only FT value of aAO1.1b A1F
Completes rigorous argument to
AnswerMarks Guidance
obtain −cos42θ+c OEAO2.1 R1
Total9
QMarking Instructions AO 
Question 8:
--- 8(a) ---
8(a) | Recalls a correct trig identity, which
could lead to a correct answer | AO1.2 | B1 | ( LHS≡)
sin2x
1+tan2 x
2sinxcosx
≡
1+tan2 x
2sinxcosx
≡
sec2 x
≡2sinxcosxcos2 x
≡2sinxcos3 x
(≡RHS )
Demonstrates a strategy for
proving the identity, eg by
converting all the terms on the LHS
to cos and sin. | AO3.1a | M1
Concludes a rigorous mathematical
argument to prove given identity
AG | AO2.1 | R1
--- 8(b) ---
8(b) | Uses identity to write integrand in
the form asin2θcos32θ | AO1.1a | M1 | 4sin4θ
∫ dθ=∫8sin2θcos32θdθ
1+tan22θ
Let u=cos2θ
du 1 du
then =−2sin2θ⇒sin2θ=−
dθ 2dθ
du
I =−4∫u3 dθ
dθ
=−4∫u3du
=−u4+c
=−cos42θ+c
Correctly writes integrand as
8sin2θcos32θ | AO1.1b | A1
Selects an appropriate method for
integrating,
e.g. substitution u =cos2θ,
or by inspection PI by sight of
cos42θ | AO3.1a | M1
k∫u3du
Obtains correctly
PI by solution in form kcos42θ, if
by inspection | AO1.1a | M1
Obtains−u4 or −cos42θOE
Only FT value of a | AO1.1b | A1F
Completes rigorous argument to
obtain −cos42θ+c OE | AO2.1 | R1
Total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac{\sin 2x}{1 + \tan^2 x} = 2\sin x \cos^3 x$
[3 marks]

\item Hence find $\int \frac{4\sin 4\theta}{1 + \tan^2 2\theta} d\theta$
[6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2018 Q8 [9]}}
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