| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Deduce related solution |
| Difficulty | Standard +0.3 This is a structured trigonometric equation question with clear scaffolding. Part (a)(i) is algebraic manipulation using tan θ = sin θ/cos θ and standard identities (3 marks guides students through the algebra). Part (a)(ii) is straightforward solving after factorization. Part (b) is a simple substitution applying previous results to θ/2. The question requires competent algebraic manipulation and understanding of trigonometric identities, but the heavy scaffolding and standard techniques make it slightly easier than average for AS-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a)(i) | Uses tanθ = identity | |
| sin𝜃𝜃 | 1.2 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Uses sin2θ + cos2θ = 1 identity | 1.2 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a)(ii) | Obtains any two solutions | |
| (AWRT) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (AWRT) | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 8(b) | Deduces that the required |
| Answer | Marks | Guidance |
|---|---|---|
| PI by ±83°or ±82° or ±240° | 2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| no further solutions. | 1.1b | A1 |
| Subtotal | 2 | |
| Question Total | 7 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a)(i) ---
8(a)(i) | Uses tanθ = identity
sin𝜃𝜃 | 1.2 | M1 | 3sinθ tanθ = 5cosθ – 2
3sinθ = 5cosθ – 2
sin𝜃𝜃
3sin2θ c = o s 5 𝜃𝜃 co s2θ – 2cosθ
3(1 – cos2θ) = 5cos2θ – 2cosθ
8cos2θ – 2cosθ – 3 = 0
(4cosθ – 3)(2cosθ + 1) = 0
cos𝜃𝜃
Uses sin2θ + cos2θ = 1 identity | 1.2 | M1
Manipulates to obtain the given
equation | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(a)(ii) ---
8(a)(ii) | Obtains any two solutions
(AWRT) | 1.1a | M1 | θ = ±41° and ±120°
Obtains all four solutions
(AWRT) | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Deduces that the required
solutions are double their
previous solutions
PI by ±83°or ±82° or ±240° | 2.2a | M1 | θ = ±41.4 and ±120°
1
2
θ = ±83°
Obtains ±83° or AWRT ±83° and
no further solutions. | 1.1b | A1
Subtotal | 2
Question Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$3\sin\theta\tan\theta = 5\cos\theta - 2$$
is equivalent to the equation
$$(4\cos\theta - 3)(2\cos\theta + 1) = 0$$
[3 marks]
\item Solve the equation
$$3\sin\theta\tan\theta = 5\cos\theta - 2$$
for $-180° \leq \theta \leq 180°$
[2 marks]
\end{enumerate}
\item Hence, deduce all the solutions of the equation
$$3\sin\left(\frac{1}{2}\theta\right)\tan\left(\frac{1}{2}\theta\right) = 5\cos\left(\frac{1}{2}\theta\right) - 2$$
for $-180° \leq \theta \leq 180°$, giving your answers to the nearest degree.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q8 [7]}}