| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Deduce related solution |
| Difficulty | Moderate -0.3 This is a straightforward trigonometric equation question requiring standard identities (tan θ = sin θ/cos θ) and routine algebraic manipulation. Part (a)(i) is guided with 1 mark, part (a)(ii) involves solving tan²θ = 4 in a given range (standard procedure), and part (b) is a direct application using the substitution 3α for θ. While it requires multiple steps and careful attention to the range in part (b), it involves only standard techniques with no novel problem-solving or insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| 4(a)(i) | Completes reasoned argument |
| Answer | Marks | Guidance |
|---|---|---|
| cosθ | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a)(ii) | Obtains tan θ = 2 or tan θ = -2 | |
| PI by one correct value for θ | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 63°, 117°, 243°, 297° | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 63°, 117°, 243°, 297° | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 4(b) | Deduces that it is necessary to |
| Answer | Marks | Guidance |
|---|---|---|
| PI by one correct value for α | 2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| values for α | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 159° | 1.1b | A1 |
| Subtotal | 3 | |
| Question 4 Total | 7 | |
| Q | Marking instructions | AO |
Question 4:
--- 4(a)(i) ---
4(a)(i) | Completes reasoned argument
to obtain tan2θ=4using
sinθ
tanθ=
cosθ | 2.1 | R1 | sinθ
tanθ=4
cosθ
tanθ×tanθ=4
tan2θ=4
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 4(a)(ii) ---
4(a)(ii) | Obtains tan θ = 2 or tan θ = -2
PI by one correct value for θ | 1.1a | M1 | tan θ = ±2
θ = 63°, 117°, 243°, 297°
Obtains any two solutions for θ
AWRT 63°, 117°, 243°, 297° | 1.1b | A1
Obtains all four solutions for θ
and no extras within
0°<θ<360°
AWRT 63°, 117°, 243°, 297° | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 4(b) ---
4(b) | Deduces that it is necessary to
divide solutions from (a) by 3
PI by one correct value for α | 2.2a | M1 | α = 21°, 39°, 81°, 99°, 141°, 159°
Obtains at least three correct
values for α | 1.1b | A1
Obtains all six solutions for α
and no extras within
0°<θ<180°
AWRT 21°, 39°, 81°, 99°, 141°,
159° | 1.1b | A1
Subtotal | 3
Question 4 Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item By using a suitable trigonometric identity, show that the equation
$$\sin \theta \tan \theta = 4 \cos \theta$$
can be written as
$$\tan^2 \theta = 4$$
[1 mark]
\item Hence solve the equation
$$\sin \theta \tan \theta = 4 \cos \theta$$
where $0^\circ < \theta < 360^\circ$
Give your answers to the nearest degree.
[3 marks]
\end{enumerate}
\item Deduce all solutions of the equation
$$\sin 3\alpha \tan 3\alpha = 4 \cos 3\alpha$$
where $0^\circ < \alpha < 180^\circ$
Give your answers to the nearest degree.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q4 [7]}}