| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Show then solve substituted equation |
| Difficulty | Standard +0.8 This is a multi-part question requiring algebraic manipulation to transform a trig equation (part a), solving a quadratic in cot (part b), and applying a compound angle substitution (part c). While the individual techniques are A-level standard, the combination of steps, the non-routine transformation to cot, and the compound angle extension make this moderately challenging—above average but not exceptionally difficult. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9\sin^2\theta + \sin 2\theta = 8\) | B1 (1.2) | Recalls \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(9\sin^2\theta + 2\sin\theta\cos\theta = 8\), uses \(\cot^2\theta + 1 = \csc^2\theta\) or \(\tan^2\theta + 1 = \sec^2\theta\); \(9 + 2\cot\theta = 8\csc^2\theta\) | M1 (1.1a) | Condone sign error |
| \(9 + 2\cot\theta = 8(\cot^2\theta + 1)\) | M1 (1.1a) | Divides through by \(\cos^2\theta\) or \(\sin^2\theta\) |
| \(8\cot^2\theta - 2\cot\theta - 1 = 0\) | R1 (2.1) | Completes rearrangement AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cot\theta = -\frac{1}{4}\) or \(\cot\theta = \frac{1}{2}\); \(\tan\theta = -4\) or \(\tan\theta = 2\) | M1 (1.1a) | PI by sight of \(2\) and \(-4\) or \(-\frac{1}{4}\) and \(\frac{1}{2}\), or two correct answers |
| \(\theta = 1.82,\ 4.96\) and \(\theta = 1.11,\ 4.25\) (two correct values) | A1 (1.1b) | Condone AWRT correct answers |
| \(\theta = 1.11, 1.82, 4.25, 4.96\) all four, no extras | A1 (1.1b) | AWRT; CAO; ignore solutions outside interval |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(2x - \frac{\pi}{4}\) equal to at least one of their solutions | M1 (3.1a) | PI by a correct answer |
| \(2x - \frac{\pi}{4} = 1.107..., 1.815...\), so \(x = 0.9, 1.3\) | A1 (1.1b) | Correct values rounded from \(0.94627...\) and \(1.300058...\); ISW once correct answers seen; CSO; Condone extra values outside of the interval |
## Question 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9\sin^2\theta + \sin 2\theta = 8$ | B1 (1.2) | Recalls $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $9\sin^2\theta + 2\sin\theta\cos\theta = 8$, uses $\cot^2\theta + 1 = \csc^2\theta$ or $\tan^2\theta + 1 = \sec^2\theta$; $9 + 2\cot\theta = 8\csc^2\theta$ | M1 (1.1a) | Condone sign error |
| $9 + 2\cot\theta = 8(\cot^2\theta + 1)$ | M1 (1.1a) | Divides through by $\cos^2\theta$ or $\sin^2\theta$ |
| $8\cot^2\theta - 2\cot\theta - 1 = 0$ | R1 (2.1) | Completes rearrangement AG |
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## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cot\theta = -\frac{1}{4}$ or $\cot\theta = \frac{1}{2}$; $\tan\theta = -4$ or $\tan\theta = 2$ | M1 (1.1a) | PI by sight of $2$ and $-4$ or $-\frac{1}{4}$ and $\frac{1}{2}$, or two correct answers |
| $\theta = 1.82,\ 4.96$ and $\theta = 1.11,\ 4.25$ (two correct values) | A1 (1.1b) | Condone AWRT correct answers |
| $\theta = 1.11, 1.82, 4.25, 4.96$ all four, no extras | A1 (1.1b) | AWRT; CAO; ignore solutions outside interval |
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $2x - \frac{\pi}{4}$ equal to at least one of their solutions | M1 (3.1a) | PI by a correct answer |
| $2x - \frac{\pi}{4} = 1.107..., 1.815...$, so $x = 0.9, 1.3$ | A1 (1.1b) | Correct values rounded from $0.94627...$ and $1.300058...$; ISW once correct answers seen; CSO; Condone extra values outside of the interval |
**Subtotal: 2 marks**
---8
\begin{enumerate}[label=(\alph*)]
\item Given that
$$9 \sin ^ { 2 } \theta + \sin 2 \theta = 8$$
show that
$$8 \cot ^ { 2 } \theta - 2 \cot \theta - 1 = 0$$
8
\item Hence, solve
$$9 \sin ^ { 2 } \theta + \sin 2 \theta = 8$$
in the interval $0 < \theta < 2 \pi$\\
Give your answers to two decimal places.\\
8
\item Solve
$$9 \sin ^ { 2 } \left( 2 x - \frac { \pi } { 4 } \right) + \sin \left( 4 x - \frac { \pi } { 2 } \right) = 8$$
in the interval $0 < x < \frac { \pi } { 2 }$\\
Give your answers to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q8 [9]}}