| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove algebraic trigonometric identity |
| Difficulty | Standard +0.3 Part (a) is a straightforward algebraic proof combining fractions and using the Pythagorean identity. Part (b) requires recognizing that cosec²θ ≥ 1 (or 2cosec²θ ≥ 2), which is standard range work. Part (c) involves solving for θ and finding cot from the given constraint. This is a well-structured multi-part question requiring standard A-level techniques without novel insight, making it slightly easier than average. |
| 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^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Recalls \(\cosec\theta = \frac{1}{\sin\theta}\); uses \(\cosec^2\theta = \frac{1}{\sin^2\theta}\) | B1 | PI by use of identity |
| Recalls \(\cos^2\theta + \sin^2\theta = 1\) | B1 | OE |
| Forms single fraction with denominator \((1-\cos\theta)(1+\cos\theta)\): \(\frac{1+\cos\theta+1-\cos\theta}{(1-\cos\theta)(1+\cos\theta)} = \frac{2}{1-\cos^2\theta}\) | M1 | OE |
| Completes to \(\frac{2}{\sin^2\theta} \equiv 2\cosec^2\theta\) | R1 | Uses \(\cos^2\theta + \sin^2\theta = 1\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Forms \(2\cosec^2\theta = A\) or \(\frac{2}{\sin^2\theta} = A\) | M1 | OE; PI by \(A \geq 2\) |
| Explains \(\cosec\theta \leq -1\) or \(\cosec\theta \geq 1\), hence \(\cosec^2\theta \geq 1\); OR \(-1 \leq \sin\theta \leq 1\) so \(\sin^2\theta \leq 1\); OR accurate sketch of \(y = \cosec^2\theta\) with 1 labelled on \(y\)-axis | E1 | Condone strict inequalities |
| Deduces \(A \geq 2\) | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses identity from (a): \(2(1+\cot^2\theta)=16\) or \(\cosec^2\theta=8\), or \(\sin^2\theta=\frac{1}{8}\) or \(\cos^2\theta=\frac{7}{8}\) | M1 | |
| \(\cot^2\theta = 7\) | A1 | PI by \(\cot\theta = \sqrt{7}\) or \(\cot\theta = -\sqrt{7}\) |
| \(\cot\theta = -\sqrt{7}\) | R1 | Deduces negative value since \(\theta\) is obtuse |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses identity from (a) to obtain \(\sin^2\theta = \frac{1}{8}\); OR rearranges \(\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=16\) to get \(\cos^2\theta=\frac{7}{8}\) | M1 | |
| \(\cos\theta = \sqrt{\frac{7}{8}}\) or \(\cos\theta = -\sqrt{\frac{7}{8}}\) | A1 | OE; must be exact |
| \(\cos\theta = -\sqrt{\frac{7}{8}}\) | R1 | Since \(\theta\) is obtuse; OE; must be exact |
## Question 8:
**Part 8(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Recalls $\cosec\theta = \frac{1}{\sin\theta}$; uses $\cosec^2\theta = \frac{1}{\sin^2\theta}$ | B1 | PI by use of identity |
| Recalls $\cos^2\theta + \sin^2\theta = 1$ | B1 | OE |
| Forms single fraction with denominator $(1-\cos\theta)(1+\cos\theta)$: $\frac{1+\cos\theta+1-\cos\theta}{(1-\cos\theta)(1+\cos\theta)} = \frac{2}{1-\cos^2\theta}$ | M1 | OE |
| Completes to $\frac{2}{\sin^2\theta} \equiv 2\cosec^2\theta$ | R1 | Uses $\cos^2\theta + \sin^2\theta = 1$; AG |
**Part 8(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Forms $2\cosec^2\theta = A$ or $\frac{2}{\sin^2\theta} = A$ | M1 | OE; PI by $A \geq 2$ |
| Explains $\cosec\theta \leq -1$ or $\cosec\theta \geq 1$, hence $\cosec^2\theta \geq 1$; OR $-1 \leq \sin\theta \leq 1$ so $\sin^2\theta \leq 1$; OR accurate sketch of $y = \cosec^2\theta$ with 1 labelled on $y$-axis | E1 | Condone strict inequalities |
| Deduces $A \geq 2$ | R1 | |
**Part 8(c) [Original paper]:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses identity from (a): $2(1+\cot^2\theta)=16$ or $\cosec^2\theta=8$, or $\sin^2\theta=\frac{1}{8}$ or $\cos^2\theta=\frac{7}{8}$ | M1 | |
| $\cot^2\theta = 7$ | A1 | PI by $\cot\theta = \sqrt{7}$ or $\cot\theta = -\sqrt{7}$ |
| $\cot\theta = -\sqrt{7}$ | R1 | Deduces negative value since $\theta$ is obtuse |
**Part 8(c) [Modified paper]:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses identity from (a) to obtain $\sin^2\theta = \frac{1}{8}$; OR rearranges $\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=16$ to get $\cos^2\theta=\frac{7}{8}$ | M1 | |
| $\cos\theta = \sqrt{\frac{7}{8}}$ or $\cos\theta = -\sqrt{\frac{7}{8}}$ | A1 | OE; must be exact |
| $\cos\theta = -\sqrt{\frac{7}{8}}$ | R1 | Since $\theta$ is obtuse; OE; must be exact |
---8
\begin{enumerate}[label=(\alph*)]
\item Given that $\cos \theta \neq \pm 1$, prove the identity
$$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } \equiv 2 \operatorname { cosec } ^ { 2 } \theta$$
8
\item Hence, find the set of values of $A$ for which the equation
$$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } = A$$
has real solutions.\\
Fully justify your answer.\\
8
\item Given that $\theta$ is obtuse and
$$\frac { 1 } { 1 - \cos \theta } + \frac { 1 } { 1 + \cos \theta } = 16$$
find the exact value of $\cot \theta$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q8 [10]}}